LoRA Probe Activations - Gate, Up, and Down Projections

All projections show: input activations ⋅ A matrix (rank-1 LoRA neuron activations)

Layer 0

GATE_PROJ

Top 16 Positive Activations
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<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nOne base of a trapez
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<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nGiven a rational number, write it
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<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet $ \mathcal{
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<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nFind the remainder when $9 \
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<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nKathy has $5$ red
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<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet's say a language $
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<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nCompute the mean molecular speed v in
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<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nTwo capacitors with capacitance values
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<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nSuppose $a,\,
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<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nJoanie takes a $\$6
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<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nIn a mathematics test number of participants
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<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nA function $f:[0
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<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet $n\geq
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<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nSquare $AIME$ has sides
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<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nConsider the following two person game.
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<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nProve that the function \[
Top 16 Negative Activations
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.003606)^2 �� (0.0036)^2 =
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(0.008)^2 ) �� sqrt(0.000013
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0*(1.03)^16 �� 6000 *1.60
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/(π*0.222)) �� sqrt( (8*8.314
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0 * (1.12)^4 �� 6000 * 1.5
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/(π*0.028)) �� sqrt( (8*8.314
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and K=1, the maximum I is ��1.31, which is less than
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so 1.03^16 �� 1.6047064
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) = 10^-2.5 �� 3.1623e-3
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0.000064) �� sqrt(0.000077
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.\n\nLinear approximation:\n\nsqrt(x + dx) �� sqrt(x) + dx/(2*sqrt(x
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0.000077) �� 0.0087746
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,150 /4975 �� Compute 4975*119
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). sqrt(7.696) �� 2.774, then multiply by
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150 / 5025 �� let's divide 6,060,
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, π * 0.222 �� 3.1416 * 0

UP_PROJ

Top 16 Positive Activations
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θ cosθ = 0\n\nFactor out sinθ:\n\nsinθ [3 + (15/
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sinθ + (15/2) sinθ cosθ = 0\n\nFactor out sinθ
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θ] = 0\n\nSolutions: sinθ =0 → ��=0, π,
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θ = 0\n\nSimplify:\n\n3 sinθ + (15/2) sinθ cos
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/dθ = (1 - ν2) sinθ cosθ / t\n\nExpress sinθ cosθ
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)=0\n\nFactor out 2i sinθ:\n\n2i sinθ[ 2 cosθ
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as 2 sintheta cos theta:\n\n3 sinθ + (15/4)*2 sinθ
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diverging.\n\nAlternatively, with substitution t = sinθ, but seems not applicable here. Wait, let
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\n\nDivide by 2i:\n\n3 sinθ + (15/4) sin2θ
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(1 - ν2) * 2 sinθ cosθ ] / [ 2 sqrt(ν
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) sinθ cosθ / t\n\nExpress sinθ cosθ in terms of t:\n\nsinθ cos
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sinθ cosθ = ?\n\nLet me express sinθ from the substitution:\n\nsinθ = sqrt( (
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θ + (15/4)*2 sinθ cosθ = 0\n\nSimplify:\n\n3
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4i sinθ cosθ -2i sinθ ( r3 +1/r3 )=0
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*2 sinθ cosθ =4i sinθ cosθ\n\nSo, rewrite the equality:\n\n4
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2θ. Then when t=ν, sinθ=0, ��=0; when t
Top 16 Negative Activations
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first problem, combining multiple sparse sets into one via tagging. For problem 2, P_bad-angel
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constructing such infinitely many positions by induction.\n\nLet me try induction. Suppose we can show there exists an increasing
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much more money Joanie will owe if the interest compounds quarterly rather than annually, we use the compound interest
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How much more money will she owe if the interest compounds quarterly than if the interest compounds annually? Express
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how much more money she will owe if the interest compounds quarterly compared to if it compounds annually. Hmm,
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infty f(x) dx$.\n\nIf I try Cauchy-Schwarz on $I$,
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<|im_start|>assistant\n<|im_start|>think\nOkay, let me try to work through these problems one by one. They
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2 more if the interest compounds quarterly than if it compounds annually.\n\nBut to make sure, let me confirm
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owe if the interest compounds quarterly than if the interest compounds annually? Express your answer as a dollar value
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<|im_start|>assistant\n<|im_start|>think\nAlright, let's try to figure out this problem. So, we have
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owe if the interest compounds quarterly compared to if it compounds annually. Hmm, I need to calculate the amount
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be of this form?\n\nAlternately, let me try to create y. Assume that y is a
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nu < 1 \). Hmm, let's try to figure out how to approach this. \n\nFirst
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/12^{1/3}=12^{-1/3}, so λ2=12
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12)^{1/3}=12^{-1/3}.\n\nThus, μ=λ
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K=1, λ=μ=12^{-1/3}\n\nCompute I=1/(4

DOWN_PROJ

Top 16 Positive Activations
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that such a point y exists, which is equidistant to all points in S with distance d/s
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constructed by translating an (assumed) existing equidistant set contained 0 and other points. Then
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orthogonality condition comes automatically from the equidistant condition. So that's interesting. Therefore,
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Therefore, if such a y exists which is equidistant to all x in S with distance d/s
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0.\n\nSo actually, if y is equidistant to x and x' with distance d/s
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the distances are equal, it's called an equidistant set. In Hilbert spaces, such sets
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the key difference is S cannot just be any equidistant set, but because the problem tells us that
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the inner product is zero. So, the orthogonality condition comes automatically from the equidistant
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.\n\nBut in our previous equation coming from the orthogonality condition in the original problem, ⟨x
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2/2 for some c, then the orthogonality condition holds.\n\nTherefore, if we can
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0.\n\nSo the point y must be equidistant from all points in S, with distance d
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fty f(x) dx <= ��A^infty (x f(x))/A dx =K
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But look back at the previous equation derived from orthogonale:\n\nFrom above, using orthogon
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orthogonale:\n\nFrom above, using orthogonality condition:\n\n||y||2 - ⟨
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^infty f dx = ��A^infty f(x) *1<= sqrt( �
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(x) *1 dx = ��0^infty f(x)*(1/(sqrt(w(x)))
Top 16 Negative Activations
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10P5. Yeah.\n\nBut perhaps another perspective: if she's selecting 5 cards from
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x' - y⟩ = 0.\n\nSo another way to think of this is that the set {
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which here is impossible. Therefore, total number of unordered pairs would be \(2^{8 - 1
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'||2 - d2)/2.\n\nBut perhaps another way is expressing this by defining various terms. Let
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65 has a new score70. But another with original60 has new65, but
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^k ordered pairs. Thus, the number of unordered pairs, accounting for swapping a and b, is
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, but as I need to compute manually... Maybe another way.\n\nAlternatively, write 1.25
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each subset has a unique complement, the number of unordered pairs is indeed 2^{k-1},
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cross-verify.\n\nWait, let’s think of another approach. For instance, imagine that we consider the
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ity is not 5, or maybe there's another way. Let's create a table further.\n\nWait
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cubic equations. So maybe let’s consider a = ��, cube roots of unity. But they have
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, non-real, not on unit circle such that ��^3=1. But also prohibited
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,22,34,...\n\nPerhaps there's another approach here. Let me reconsider.\n\nAn idea in
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. Hence, no.\n\n Alternatively, if a = �� k, b=ω^2 k,
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Wait, that seems familiar but let me verify using another approach.\n\nSince 1.25 *
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variables are scaled.\n\nAlternatively, taking a complex number ��, non-real, not on unit circle such

Layer 1

GATE_PROJ

Top 16 Positive Activations
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you add to S, you must exclude only finitely many others (i.e., n +s
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. Since each existing a_j rules out only finitely many (about2sqrt(a_{i+
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if and only if it is polynomial-time Turing reducible to a sparse set. So that's the
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able in poly-time, there's no nondeterminism. Since the angel string is computable in poly
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the midpoints of the legs divides the trapezoid into two regions whose areas are in the ratio
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, Grundy numbers aren't showing a clear periodicity. Therefore, it's going to be challenging to
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the trapezoid into two smaller trapezoids? Each with height h/2, if
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third is 999 (three 9s), etc., up to a number with 9
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impossible. Hence under any circumstances,given these constraintsthe answer to part(b) is that there is
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angel}} \) does not provide the nondeterminism inherent in \( \textbf{NP} \
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at some point, larger square numbers alter the periodicity. Let me recast the previous positions and see
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The line at height k creates a smaller trapezoid with bases a and x and height k.
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oly (canonical example given was unary languages including undecidable ones), then L ∈ P_angel is
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. It is known that P/poly contains undecidable languages (as the advice can be uncomput
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the midpoints of the legs divides the trapezoid into two regions with areas in the ratio
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a single string, and find a collision-resistant hash of α_n. Not sure.\n\nWait, think about
Top 16 Negative Activations
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)=sqrt( J )= /2.\n\nCombined: I<= (2K)^{1/
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as t goes from ν to1. Then,\n\nIntegral f(nu) = ��_{0}
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/ν. Let's plug in:\n\nThen, integral from t=ν to t=1 of (
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d2) / 2 = 0.\n\nCompare to the above line:\n\nFrom the sum of norms
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t2) )\n\nTherefore, the integral becomes:\n\nIntegral from ν to 1 of (dt / ν
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75 m/s. Which is correct.\n\nSo comparing that with Rn-222. So
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t^2) ) ] dt\n\nThen, integrating from t=ν to1, the integral f
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Let me see. Let's try substitution.\n\nThe integral is from x = 1 to x =
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00001936\n\nSo combining all:\n\n1.5625 +
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/9)c^6 /d^3\n\nCompare LHS and RHS: LHS = c^
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the 1/ν and ν cancels:\n\nIntegral from ν to 1 of dt / sqrt(
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3/2} } ] d theta\n\nBut integrating by parts perhaps to express this. Alternatively, there
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Second term: (1 - t2)\n\nSo denominator becomes sqrt( ( (t2 - ν2
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x'|| squared / 2. ).\n\nThus, matching one of the components:\n\nSo, substituting the
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. Rounded to the nearest cent.\n\nSo, if compounded annually, Joanie will owe $9,4
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Since |ω|=1. Hence, no.\n\n Alternatively, if a = �� k, b=

UP_PROJ

Top 16 Positive Activations
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as \boxed{169 \text{ m/s}}.\n\nHowever, depending on source of problem
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\sqrt{\frac{8RT}{\pi M}}\n\]\n\nwhere:\n- \( R \)
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8 * R * T) / (π * M))\n\nR = 8.314 J
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\) is the temperature in Kelvin,\n- \( M \) is the molar mass of radon
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can be represented by a tridiagonal matrix \( M \) where each row has 1s on
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\} \) and a polynomial-time machine \( M \). By the Meyer's theorem equivalence between \(
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i \leq k \). The machine \( M \) with oracle \( S \) constructs the
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.576\n\nDenominator: π * M = π * 0.222 �
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0*second + 1*third row M^3 = [0,1,0]
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\( \alpha_n \), and the machine \( M \) queries \( S_L \) for the
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,950)=167.2 m/s. So around 167 m/s
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0]]\n\nThen, M^4 = M * M^3:\n\nFirst row: 0*M^
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753 ) �� 475 m/s. Which is correct.\n\nSo comparing that with
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624; sqrt ��169 m/s.\n\nBut no one knows. Wait, but
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2) �� 168.5 m/s\n\nSo approximately 168.5
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C1 / C1 + dC2 / C2 - (dC1 + dC2
Top 16 Negative Activations
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or incomplete elliptic integral. Let's check.\n\nIndeed, in the integral:\n\n∫_{ν}^
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with countably or uncountably many vertices.\n\nIndeed, in an infinite-dimensional Hilbert space, one
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would be sqrt(2) times the radius.\n\nIndeed, if you take an orthonormal basis in
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1,3), possibly the only possible pair.\n\nThus, after all this deduction my best guess is (
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there exists a point y common in all spheres.\n\nThus, applying the finite intersection property:\n\n1. For
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this checks out.\n\nSo all steps are correct.\n\nThus, conclusion: at 298 K,
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impossible averages, leading no N satisfies the given.\n\nThus:\n\n(a) 12, 24
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9 mod125. So that matches.\n\nThus, if we follow the steps, the result should
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fine. Good.\n\nNow modulus125:\n\nAgain, Product is(10^k -1
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each new losing position avoids previous differences with squares.\n\nThus, we conclude there are infinitely many initial situations where
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to 4, there are two color sequences.\n\nThus, the number of color sequences in Case 2
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1). Therefore, the entire expression is negative.\n\nThus, f’(nu) <0 for all
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is prohibited. Thus, this approach is invalid.\n\nOkay, this is getting too tangled. Let's maybe
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linear function with support on finite interval gives optimal I≈1.31 (so I^3≈
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speed only depends on temperature and molar mass.\n\nThus, if taking STP as 0°C,
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real. which is prohibited. So again invalid.\n\nThus such setup leads to real c, which is prohibited

DOWN_PROJ

Top 16 Positive Activations
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2.6% total spread. But since normally uncertainty is defined as ± half the total spread,
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if exactly one neighbor is 1 (counted normally), but consider two neighbors: "exactly one
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is scaling-invariant. Therefore, without loss of generality, we can normalize one of the integrals
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sphere centered at y. Alternatively, without loss of generality, perhaps we can assume y is at the
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J and K to single variable.\n\nWithout loss of generality, let’s set J=1 and K
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.\n<|im_start|>user\nIn a mathematics test number of participants is $N < 40$ .
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are\nthe following: \nThe average of all participants is $66$ , that of the
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(number of pebbles) where no matter what the first player does, the second player can respond
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initial positions where the second player can win no matter what, this implies that. Hence, the conclusion is
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user\nCompute the mean molecular speed v in the heavy gas radon (Rn) in m/s
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matrix M is nilpotent, then no matter what initial state, the system evolves to zero in GF
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. Not exactly matching here.\n\nAlternatively, consider hypergeometric substitutions.\n\nWait, but maybe that's getting
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are increased by 5, it might cause some participants who previously failed to now pass. Therefore, the
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65. Originally, the average score of all participants is 66. The average of those who
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We are given:\n\n1. The average of all participants is 66. Therefore, the total score
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the number of repeaters as R. Since all participants are either promoted or repeaters, N = P
Top 16 Negative Activations
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Altizio, which suggests there is a clever solution instead of heavy computation.\n\nRecall that from abc =
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, actually, maybe my initial analysis with arithmetic (instead of mod2) was flawed.\n\nWait, no
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in some contexts, STP is 1 bar instead of 1 atm, but regardless, the formula
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1. But problem states75 and59 instead, which is inconsistent. Hence, my previous assumption
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Alternatively, if we take 273 K instead of 298 K, let's compute
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6! example. For 6!, which had instead three distinct prime factors (2,3,5
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that p and q are real. Maybe considering algebra instead of analysis. Suppose we note that given p and
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^8 as accumulating step by step.\n\nBut perhaps instead of using exponentiation, compute sequentially:\n\n1.
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maybe the problem uses R=8.31 instead of 8.314. Let me
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process is getting a bit convoluted. Maybe instead test the inequality for a specific function and see if
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0, and check over a smaller case. Suppose instead that Kathy had 1 red and 1 green
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Let me check with a smaller factorial.\n\nSuppose instead of 20!, take a simpler case.
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Wait, but log n is variable length. If instead, we represent n with O(log n) bits
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the mutual distances between points are all d, this might impose that the spheres intersect.\n\nPerhaps we can use
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of the initial state, while for others, it might always die out. The challenge is to characterize the
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.88% when rounded. Alternatively, it might be presented as 0.877%,

Layer 2

GATE_PROJ

Top 16 Positive Activations
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= |b|^2*(|b|^4 a ) ) = |b|^6 a → Therefore
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1 / sqrt( (x^2 - a^2)(b^2 - x^2
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) ]3 +3 [sqrt(2 K a )]^2 (a/2 ) +3
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/2 ) +3 (sqrt(2 K a ))(a2/4 ) +a3
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8≡3 mod5. Is 18 a winning or losing? Not sure, since previous P
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2.\n\nNeed to show ( sqrt(2 K a ) +a /2 )3 ≤8 K
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Expand S3:\n\n= [sqrt(2 K a ) ]3 +3 [sqrt(2 K
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contradicting the idea that all multiples of5 are losing. Therefore, there must be a mistake in
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c} + \overline{b^2 a} + \overline{c^2 b
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2.\n\nSo I <= sqrt( 2 K a ) + a/2.\n\nNeed to show (
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= (a^2 c + b^2 a +c^2 b)/(abc)= a^
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1/a3\n\nTerm6: 1/( a3 b3 )\n\nBut this seems like each term
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a/2 ) +3 sqrt(2 K a)(a/2 )2 + (a/
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,b>0,\n\nI^3 <= [(4 a^{-1}) J + (2^{-1}
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} [1 / sqrt( (t2 - a2)(b2 - t2) ) ]
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a is (b*(a + b) - a*b)/(a + b)^2) = (
Top 16 Negative Activations
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+ ε}^1 ... dt\n\nThen differentiating would give:\n\n- F(ν, ν)
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25\n\nMultiply through by y to eliminate the denominator:\n\ny(y + 100) -
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E(k)/nu ] <0, since the denominator is 1 -nu2>0 in the
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, the integral is well-defined. But when we differentiate f(ν), there's a delta-like contribution
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{2^k} \) for some non-negative integer \( k \).\n\n**But Wait:**\n
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1 = 0 \).\n\nSo, the denominator becomes sqrt( ( (1 - ν2)/
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So the integral is convergent, but when we differentiate, the Leibniz rule gives a problematic
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boxed{2^k}\n\]\n\nfor non-negative integers \( k \).<|im_end|>
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)3 ≤8 K a2.\n\nLet’s denote that expression. Let me compute:\n\nLet S=s
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s <=n}\n\nSo mex is the smallest non-negative integer not in the set.\n\nCalculating Grundy
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frac{K}{A}.\n\]\n\nWe then minimize the right-hand side by choosing \( A \).
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298\)\n - Compute the denominator: \(\pi \times 0.2
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a2 z )=1, which holds.\n\nRegardless, the significance of expressing each bracket as t +
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, the mol unit cancels, then kg in denominator; thus overall (kg·m2)/(s
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(x). Hmm.\n\nAlternatively, recognize that via Homogeneity: The inequality is scaling invariant. If we replace
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on the unit circle, their logs would have non-zero real parts. But I don't see how that

UP_PROJ

Top 16 Positive Activations
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We need to solve for x in:\n\nx �� 109 mod125\n\nx
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��5 mod8.\n\nTherefore, P �� 5 mod8.\n\nAlright, now onto modulo
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3 mod4:\n\nPossible moves:\n\nIf n ��0 mod4: subtract0 or1.\n\n subtract
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;0 -1=3;\n\nIf n ��1 mod4: subtract1 or0.\n\nn
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P mod 8.\n\nSince 10 �� 2 mod 8, so 10
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9 mod125.\n\nTherefore, P ��109 mod125.\n\nSo summary
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109 mod125\n\nx ��5 mod8\n\nLet x =125
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8 are coprime, m must be ��0 mod8. Therefore, m=8n
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find when (2^k -1) �� 0 mod 8, which would make the
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1*3*(-1) = -3 ��5 mod8.\n\nTherefore, P ��
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, so (2^k -1) �� -1 mod8.\n\nTherefore, for k=
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, (10^k -1) �� -1 mod125 for k≥3
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�5 mod8, hence 5m ��0 mod8, leading to m≡0 mod
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etc. So indeed, these numbers are n ��0 or 2 mod5. Wait, let
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for k ≥3, 2^k �� 0 mod8, so (2^k
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mod8, 5m + 5 ��5 mod8, hence 5m ��
Top 16 Negative Activations
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56*(N/3)-42*( (N/3) -x).\n\nSimplify:\n\n
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) = ��_{ν}^1 ( (t2 - ν2)^{-1/2
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since 8 is even). However, in many applications, rounded up. There's conflict here. But
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}^1 [ ��/∂ν ( (t2 - ν2)^{-1/2
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)) and the side at x=10 (from (10,0) to (10
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/3) + x ) +59*( (N/3) - x ) =71
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00^2) )^2 + ( (15*2000^2)/(
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the 2*0.004 (from the voltage), gives for energy:\n\nsqrt( (
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*(c/d) - 2c d*( (c/d)^2 /2 ) + d^
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= ��[0 to10] ( (10/h x) + (10 -
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^3 b^3 ) = 1/( (a b)^3 )\n\nSo replacing each term:\n\n
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integrals and use some sort of dual Hölder applications?\n\nWait, let's see. Let's denote
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= 1/ν of dx over sqrt( (x2 -1)(1 - ν2x
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0.028)) �� sqrt( (8*8.314*29
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is:\n\n∫[0 to10] [ (5/h x +10 -50/h
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= ��[0 to10] [ (5/h x +10 -50/h

DOWN_PROJ

Top 16 Positive Activations
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integral of the second kind. But I might be misremembering.\n\nAlternatively, check relations of derivatives.
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b) the meaning of passmark might have been misconsidered.\n\nWait, pass mark is fixed at
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<n, i, b>, where the i-th bit of α_n is b. But in order
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the triangle must be mostly inside the square but partially overlapping, with some part outside? Wait, but if
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't compatible (i.e., containing 0 and overlapping points not constructed from orthogonal vectors).\n\nBut in reality
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(10,10). Then, the overlapping region is the part of this triangle that's inside
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0), (10,10). The overlapping area between this triangle and the square requires calculating the
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square area is 100. So the overlapping area is 80. Therefore, the triangle
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.\n\nBut this is the overlapping area. Because The overlapping area is inside the square and inside the triangle,
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) and apex G outside the square. Then the overlapping area between the triangle and square is a trape
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0,10 -50/h). The overlapping area between the triangle and the square would then be
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50/h). Then connecting these points, the overlapping shape would be a trapezoid with vertices
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=0 and x=10.\n\nBut the overlapping region is the part of the triangle GE to GM
Rollout 9
(10,10). So the area overlapping is inside square and inside triangle.\n\nThus, the
Rollout 9
=10.\n\nSo in other words, the overlapping region is between x=0 and x=1
Rollout 9
so triangle is within. But then how come the overlapping area is 80 then? Because that that
Top 16 Negative Activations
Rollout 10
called losing positions or P-positions in combinatorial game theory. The problem states there are infinitely many
Rollout 8
only concerned with the colors. That is, the actual arrangement of colors in the sequence. So if in
Rollout 12
* 1, which is -∞, the actual result may be finite because the other part (the
Rollout 6
bin(n) concatenated with α_n, then the actual bits of α_n can be encoded in the oracle
Rollout 15
GF(2) model. However, in the actual system, it's equivalent to computing x_{t
Rollout 15
2) reduces components each time, but in the actual system, since each step's is dependent only on
Rollout 8
and green, as well as the permutations of the actual cards.\n\nBut maybe in terms of generating the happy
Rollout 3
but I might have to verify which one is the actual mean speed. Let me think.\n\nMean speed,
Rollout 7
sense. If the scores are increased, then the actual marks that count are the new ones. But the
Rollout 15
GF(2) model is insufficiently modeling the actual system. Hence, perhaps equivalence is not correct.\n\n
Rollout 15
when you track based on rules (lamp is ON tomorrow only if exactly one neighbor is ON today), irrespective
Rollout 2
000. But from the problem, rather than factorizing each term, might use modulus10
Rollout 3
. In gas experiments, diffusion rates, Graham's law, typically at STP. But the molar
Rollout 8
arrangements with two blocks".\n\nAlternatively, in combinatorics, we can model it for a mixed color layout
Rollout 11
. So p would be a real number in this setup. Similarly, check if q is real. Compute
Rollout 10
or similar. It's a result in combinatorics that you can have infinite sets where no two elements

Layer 3

GATE_PROJ

Top 16 Positive Activations
Rollout 0
is \(\boxed{128}\).<|im_end|>
Rollout 10
where the second player can force a win.}<|im_end|>
Rollout 1
y \text{ exists as described}}\n\]<|im_end|>
Rollout 9
\) is \(\boxed{25}\).<|im_end|>
Rollout 5
[\n\boxed{181}\n\]<|im_end|>
Rollout 6
) to decide \( x \in L \).<|im_end|>
Rollout 2
is \(\boxed{109}\).<|im_end|>
Rollout 11
boxed{(-1, 3)}\).<|im_end|>
Top 16 Negative Activations
Rollout 1
could be uncountable. Hmm.\n\nBut let's think: Suppose that S is a set where all
Rollout 11
't 1, maybe log variables?\n\nBut let's assume for simplicity that all variables are real? Wait
Rollout 8
n=126.\n\nWait, but let's confirm. Wait, when you sum Case1 and
Rollout 9
GE and GM) with base EM.\n\nSo let's calculate coordinates.\n\nGiven:\n\nE is (10
Rollout 10
3, resulting in an infinite sequence.\n\nBut let's verify this further. Next, n=15
Rollout 4
0.87%.\n\nBut wait, let's go step by step for verification once more.\n\nFirst
Rollout 9
+ (10 - y)^2\n\nCanceling h^2:\n\ny^2 = (1
Rollout 12
the integral or doing another substitution.\n\nWait, let's look again at the integrand in the transformed integral
Rollout 12
this term doesn't actually contribute?\n\nWait, let's recall that when we use differentiation under the integral sign
Rollout 4
15 )^2\n\nHmm, let's compute these step by step.\n\nFirst, compute the
Rollout 1
set, perhaps this intersection is possible.\n\nBut let's see the distance between any two centers (which are
Rollout 13
643736\n\nHmm, let's break it down:\n\n0.0047
Rollout 4
because the voltage is squared. \n\nWait, let's check the formula. If a quantity Q is proportional
Rollout 12
to t, but not sure.\n\nWait, let's attempt to compute the derivative f’(ν).
Rollout 15
. Wait, maybe a mistake.\n\nWait, let's think again. Let's map t=0:
Rollout 9
distance from G to EM.\n\nWait, maybe visualizing coordinates would help. Let's place EM as the

UP_PROJ

Top 16 Positive Activations
Rollout 9
common to triangle $GEM$ and square $AIME$ is $80$ square units.
Rollout 12
) \cdot b'(\nu) - F(a(\nu), \nu) \cdot a'
Rollout 7
56N/3\n\nFactor x:\n\nx(A -54) +18N =5
Rollout 4
)/(a + b)^2) = (b(a + b - a)) / (a + b
Rollout 9
common to triangle \(GEM\) and square \(AIME\) is 80 square units. We
Rollout 9
a helpful mathematics assistant.\n<|im_start|>user\nSquare $AIME$ has sides of length $10$
Rollout 7
=56N/3\n\nThus:\n\nx(A -42) +14N=5
Rollout 12
{d}{d\nu} \int_{a(\nu)}^{b(\nu)} F(x
Rollout 7
=56N/3\n\nTherefore:\n\nx(A -54) =56N/3
Rollout 14
2\n\nDivide (1) by (2):\n\n(λ2 μ)/(λ μ2)= λ
Rollout 7
x=0\n\n→-3P +x(A -79)=0\n\nBut P=2
Rollout 1
x - y⟩ = d2 / 2,\n\nwhich is ⟨x, x⟩ - 2
Rollout 1
||y||2 = (d2)/2,\n\n||x2||2 - 2⟨x
Rollout 9
boxed{25}\n\n<|im_start|>answer\nSquare \(AIME\) has sides of length 10 units
Rollout 1
:\n\n⟨x, t - x⟩ = C,\n\n⟨x', t - x'⟩ = C
Rollout 7
/3)\n\nWhich simplifies similar to part(a):\n\nA*x +42*N/3 -4
Top 16 Negative Activations
Rollout 12
. Not exactly matching here.\n\nAlternatively, consider hypergeometric substitutions.\n\nWait, but maybe that's getting
Rollout 15
would have a single 1 on the first superdiagonal and first subdiagonal.\n\nWait, example
Rollout 10
) were computed for several positions, showing non-periodic but structured behavior.\n\n4. **Constructing Infinite
Rollout 15
matrix with 1s on the sub and superdiagonals.\n\nFor instance, if you have this
Rollout 10
we can find a different pattern or a non-periodic structure.\n\nStarting from n=0 (L),
Rollout 10
5, the Grundy numbers may be non-periodic.\n\nBut since manual calculation is tedious, let's
Rollout 15
M has 1s on the super and subdiagonal:\n\nM = [[0,1],\n\n
Rollout 2
09. But wait, but I feel like in this case, the product of all these numbers,
Rollout 15
agonal with 1s on the super and subdiagonals, and 0s elsewhere. The
Rollout 15
1 on the first superdiagonal and first subdiagonal.\n\nWait, example for n=4:\n\n
Rollout 15
it has 1s on the super and subdiagonal and 0s elsewhere. For example,
Rollout 15
row has 1s on the super and subdiagonal for internal lamps, and a single 1
Rollout 10
values) were computed for several positions, showing non-periodic but structured behavior.\n\n4. **Constructing
Rollout 0
\(a = b\). So all coprime pairs with \(a \times b = 2
Rollout 1
of S and construct y accordingly.\n\nBut since non-separable Hilbert spaces have orthonormal bases of
Rollout 1
the quadratic form is affine.\n\nBut this seems non-trivial, but perhaps specific to the set S.\n\n

DOWN_PROJ

Top 16 Positive Activations
Rollout 5
[\n\boxed{181}\n\]<|im_end|>
Rollout 13
{187.12}\n\]<|im_end|>
Rollout 1
y \text{ exists as described}}\n\]<|im_end|>
Rollout 8
[\n\boxed{157}\n\]<|im_end|>
Rollout 0
is \(\boxed{128}\).<|im_end|>
Rollout 4
{0.88\%}\n\]<|im_end|>
Rollout 7
. Perhaps some students had their original score so that despite unrealistic averages, their scores sum up to needed.
Rollout 9
+ (10 - y)^2\n\nCanceling h^2:\n\ny^2 = (1
Rollout 12
on } (0, 1)}\n\]<|im_end|>
Rollout 10
where the second player can force a win.}<|im_end|>
Rollout 1
a shift and scaling.\n\nSumming up, then despite the difficulty with coordinate system, using the properties of
Rollout 14
) \, dx \right) }\n\]<|im_end|>
Rollout 11
line{c} = 1/(\overline{ab})). Therefore, maybe expressions can be
Rollout 15
any tridiagonal matrix, particularly symmetric ones, eigenvalues are real. However over GF(2),
Rollout 11
= 1/(ab). Thus, \overline{c} = 1/(\overline
Rollout 11
overline{a b}. Therefore, \overline{a b} = 1/(ab).
Top 16 Negative Activations
Rollout 1
into equations involving t. Let me see.\n\nDenote t = 2y. Then y = t
Rollout 11
not on unit circle either.\n\nAdditionally, (denote S = a + b + c, T =
Rollout 11
of the form t +1/t.\n\nThus, denote t1 =a, t2= a z
Rollout 5
system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nOne base of a trapezoid is
Rollout 0
system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nGiven a rational number, write it as a
Rollout 1
system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet $ \mathcal{H}$
Rollout 4
system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nTwo capacitors with capacitance values $C
Rollout 6
system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet's say a language $L \
Rollout 2
system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nFind the remainder when $9 \times
Rollout 9
system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nSquare $AIME$ has sides of length
Rollout 8
system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nKathy has $5$ red cards and
Rollout 7
system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nIn a mathematics test number of participants is
Rollout 3
system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nCompute the mean molecular speed v in the heavy
Rollout 10
system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nConsider the following two person game. A number
Rollout 11
system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nSuppose $a,\,b,$
Rollout 12
system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nProve that the function \[ f(\

Layer 4

GATE_PROJ

Top 16 Positive Activations
Rollout 1
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 0
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 3
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 4
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 2
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 6
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 7
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 8
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 9
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 10
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 11
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 12
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 13
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 14
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 15
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 5
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Top 16 Negative Activations
Rollout 1
Compare to the above line:\n\nFrom the sum of norms, we have:\n\n-⟨x + x',
Rollout 11
perhaps keeping variables as they are and using symmetries or equations from p and q.\n\nGiven that both
Rollout 9
this first. \n\nLet me draw the square AIME. Since it's a square, all sides are
Rollout 9
of overlap between triangle GEM and the square AIME is 80.\n\nBut how does the triangle
Rollout 1
, if we have a set of vectors with equal norms and their differences have equal norms, this often relates
Rollout 1
, S' was constructed by translating an (assumed) existing equidistant set contained 0 and
Rollout 9
the common area between triangle GEM and square AIME is 80. Since the square is from
Rollout 12
theta2 ). Hmm, this is not the standard form. However, perhaps with substitution.\n\nAlternatively, substitute
Rollout 9
the area common to triangle GEM and square AIME is 80 square units. The square area
Rollout 12
2) ) ] dx\n\nWhich is a standard form, an elliptic integral. So our integral f
Rollout 3
222). However, depending on the isotope. Natural radon is mainly Rn-2
Rollout 10
Since, even though the Grundy numbers seem aperiodic, the fact that subtracting squares allows jumps
Rollout 10
, even though the Grundy numbers seem aperiodic, the fact that subtracting squares allows jumps might
Rollout 5
length x can be found using the formula which is sort of a weighted average. Let me recall that if
Rollout 11
. It's rotationally symmetric, if we permute a→b→c→a. Similarly,
Rollout 3
-219. But the most common isotope is Rn-222, which has

UP_PROJ

Top 16 Positive Activations
Rollout 7
$47$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I'm trying to solve this
Rollout 6
$ . [/*]\n<|im_start|>assistant\n<|im_start|>think\nOkay, let me try to work through these
Rollout 1
mathcal{H}$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to show that there
Rollout 11
.\n\n*David Altizio*\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I have this problem here where
Rollout 12
[P. Turan]\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to prove that the
Rollout 14
dx \right) $$ \n<|im_start|>assistant\n<|im_start|>think\nOkay, so I have to prove this inequality
Rollout 10
matter how his opponent plays.\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to prove that in
Rollout 2
1000$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I have this problem here:
Rollout 9
$\triangle GEM$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to solve this geometry
Rollout 3
n) in m/s\n<|im_start|>assistant\n<|im_start|>think\nOkay, I need to calculate the mean molecular
Rollout 15
be off after some time?\n<|im_start|>assistant\n<|im_start|>think\nAlright, let's try to figure out this
Rollout 13
value to the nearest cent.\n<|im_start|>assistant\n<|im_start|>think\nOkay, so Joanie took a $6
Rollout 4
this combination of capacitors?\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to find the percentage
Rollout 5
/100$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. I need to
Rollout 8
$m + n$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. We need to
Rollout 0
$ be the resulting product?\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I need to figure out how
Top 16 Negative Activations
Rollout 15
polynomial is x^n.\n\nBut for any tridiagonal matrix, particularly symmetric ones, eigenvalues are real
Rollout 15
Such matrices are called Jacobi matrices or tridiagonal matrices. Nilpotency can be discussed via their
Rollout 15
- The system can be represented by a tridiagonal matrix \( M \) where each row has
Rollout 15
over GF(2), M is a tridiagonal matrix with 1s on the sub and super
Rollout 15
on the first superdiagonal and first subdiagonal.\n\nWait, example for n=4:\n\nM
Rollout 9
y-coordinate) and M (10 in y-coordinate). That makes sense for an isosceles
Rollout 9
5, halfway between E (0 in y-coordinate) and M (10 in y-coordinate).
Rollout 15
has 1s on the super and subdiagonal:\n\nM = [[0,1],\n\n [
Rollout 11
b, c are inverses of the conjugate variables? Not sure. Maybe if |a| ≠
Rollout 11
+\dfrac ca$ .\n\end{tabular}\nGiven that both $p$ and
Rollout 15
considering the path graph's adjacency matrix is tridiagonal with 1s on super and subdiagonal
Rollout 15
have a single 1 on the first superdiagonal and first subdiagonal.\n\nWait, example for
Rollout 15
neighbor). The problem asks for which n the cellular automaton is nilpotent. It's equivalent to
Rollout 15
has 1s on the super and subdiagonal for internal lamps, and a single 1 for
Rollout 13
logarithms. Oh, maybe better to compute step by step.\n\nWe can also use the formula for compound
Rollout 15
linear over GF(2). For a linear cellular automaton over a path graph with rules as defined (

DOWN_PROJ

Top 16 Positive Activations
Rollout 11
p=(a+b+c)+\left(\dfrac 1a+\dfrac 1b+\
Rollout 10
other relation in P-positions. However, stepping back, the answer likely is showing that multiples of
Rollout 11
} + e^{iθ}\n\nRearranging terms:\n\nLeft side: 4 e^{i
Rollout 11
but not obviously helpful.\n\nWait, let's step back and revisit the key information given. All relevant variables
Rollout 1
/ 2 = 0.\n\nRearranging:\n\n||y||2 - ⟨x + x
Rollout 1
is a constant depending only on t.\n\nLet me rearrange this as:\n\n⟨x, t - x
Rollout 14
} $$\n\nWait, no. Let me step back. Hölder states that ��g h k
Rollout 9
,0).\n\nWait, wait. Let's step back. The triangle GEM is outside except for the
Rollout 10
force a win by leveraging the structured analysis and inductive construction of losing positions.\n\n\boxed{There are
Rollout 1
) / 2 = 0.\n\nRearranging:\n\n||y||2 - ⟨x +
Rollout 14
bit tricky. Maybe express this as:\n\nRearranged,\n\n$$\nf(x) = a -
Rollout 11
θ} + e^{iθ}\n\nRearranging terms:\n\nLeft side: 4 e^{
Rollout 1
||2 = d2/2.\n\nRearranged:\n\n2⟨xi, y⟩ = ||
Rollout 1
⟩ - d2 / 2.\n\nSo, rearranged:\n\n⟨x, y⟩ = (⟨
Rollout 10
a losing position.\n\nHence, according to Hindman's theorem or Folkman's theorem, though probably
Rollout 1
a constant depending only on t.\n\nLet me rearrange this as:\n\n⟨x, t - x⟩
Top 16 Negative Activations
Rollout 11
so c = 1/a = b, and then b = 1/c = 1/b.
Rollout 11
is if their expressions in p and q are a result of symmetric arrangements.\n\nAlternatively, since abc =1
Rollout 15
2^1 -1 =1, and then maybe 3, 7, 15
Rollout 15
goes to [1,0,1] and then to [0,0,0], i.e
Rollout 14
$I \cdot I \cdot I$, and then trying to bound each $I$ by different integr
Rollout 3
So approx 161 m/s.\n\nSo depending on the temperature assumption, the answer would be different
Rollout 9
y=0 to y=10. So depending on the value of h, the triangle might be
Rollout 7
59$ .\n(a) Find all possible values ​​of $N$ .\n(b) Find all
Rollout 3
161.4 m/s.\n\nSo, depending on the temperature assumed (273 K vs
Rollout 10
where each n_i is a losing position, and then show if we can construct n_{i+1
Rollout 10
0,22,34,...(OEIS A017101?), but verification
Rollout 3
weight is about (222). However, depending on the isotope. Natural radon is mainly
Rollout 0
assigned to a must go completely into a) and then b is 20! / a, with
Rollout 3
9 \text{ m/s}}.\n\nHowever, depending on source of problem, maybe requires two sig fig
Rollout 5
trapezoid into two equal areas, and then find the greatest integer not exceeding x2/1
Rollout 1
three spheres is likely a finite set or empty. Depending on the configuration. For regular simplex, in

Layer 5

GATE_PROJ

Top 16 Positive Activations
Rollout 5
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 0
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 1
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 2
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 3
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 4
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 11
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 6
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 7
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 8
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 9
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 10
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 12
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 13
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 14
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 15
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Top 16 Negative Activations
Rollout 6
n + n^k �� m? Unlikely. For example, m is dominated by n^
Rollout 1
vectors could form an orthogonal set with each vector having norm d / sqrt(2}, so that the distance
Rollout 11
where d is some variable. But likely overcomplicating.\n\nAlternatively, assume a = d, b
Rollout 6
, no, because the machine M doesn't have prior knowledge of α_n. The machine M needs to
Rollout 1
must be an orthogonal set, with each vector having norm d / sqrt(2). Then scaling by sqrt
Rollout 0
will satisfy a < b, even for coprime divisors. Wait, but in reality, since
Rollout 1
finite-dimensional space, in order to have equidistant points, they must form a regular simplex, which
Rollout 0
*b=N!:\n\nSince a and b are coprime.\n\nHence equivalence between choices over subsets of
Rollout 1
y is the origin. Then, all x have norm d / sqrt(2)), and are pairwise orthogonal
Rollout 0
a and b? But since they must be coprime, it's either assigned entirely to a or
Rollout 2
10 and 125 are coprime? 10 and 125
Rollout 0
assignments will satisfy a < b, even for coprime divisors. Wait, but in reality,
Rollout 1
that { x - y } is orthogonal system with norm d / sqrt(2).\n\nFrom Rudin or
Rollout 1
the set {x - y} is orthogonal with norms d/sqrt(2).\n\nAlternatively, is this
Rollout 0
. When we say that a and b are coprime with a*b=20!, then a
Rollout 0
in general, 2^{k} ordered coprime pairs (a,b), and because n!

UP_PROJ

Top 16 Positive Activations
Rollout 12
period depending on a and b. Wait, more precisely, for the integral in the form:\n\n∫_
Rollout 1
where each pair is separated by d, perhaps S must lie on such a quadratic manifold. But unless I
Rollout 14
space. Maybe try expanding on a basis. Hmmm...\n\nWait I just had another thought. Maybe use
Rollout 4
0.88%, or maybe the answer is exact at 0.87%.\n\nBut wait
Rollout 6
poly-time machine to access the bits. Another approach might be to store α_n in the oracle using a
Rollout 10
greater than k2 for k >2. Might not help.\n\nAnother angle: If n is such that
Rollout 14
this equality holds for arbitrary J and K. Maybe impossible. Therefore, move away from scaling.\n\nAlternatively,
Rollout 12
Jacobi elliptic function. Wait, maybe not necessary. Alternatively, use a trigonometric substitution.\n\n
Rollout 3
298 K (25°C), answer is approximately 168.5 m/s.\n\n
Rollout 14
and a <1 for收敛 at zero. Can’t both. Thus integral over(0, ��
Rollout 14
matches the ratio here. Hmmm.\n\nWait a minute, perhaps I made a miscalculation in
Rollout 7
=98.\n\nBut that's impossible as A must be <=64 (since original repeaters with
Rollout 1
(2) } is not convex. Oh, right. The sphere is not convex. So perhaps closed
Rollout 3
168.5\n\nThus, answer is approximately 168.5 m/s.
Rollout 12
* infty, which is more nuanced. Hmmm. Let's see.\n\nTake x =1 -
Rollout 2
the remainder is109.\n\nTherefore, answer:109.\n\nBut let me recheck all
Top 16 Negative Activations
Rollout 8
$m + n$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. We need to
Rollout 10
matter how his opponent plays.\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to prove that in
Rollout 11
.\n\n*David Altizio*\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I have this problem here where
Rollout 3
n) in m/s\n<|im_start|>assistant\n<|im_start|>think\nOkay, I need to calculate the mean molecular
Rollout 2
1000$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I have this problem here:
Rollout 13
value to the nearest cent.\n<|im_start|>assistant\n<|im_start|>think\nOkay, so Joanie took a $6
Rollout 12
[P. Turan]\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to prove that the
Rollout 0
$ be the resulting product?\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I need to figure out how
Rollout 4
this combination of capacitors?\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to find the percentage
Rollout 9
0 units. Isosceles triangle \(GEM\) has base \(EM\), and the area
Rollout 9
$\triangle GEM$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to solve this geometry
Rollout 7
$47$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I'm trying to solve this
Rollout 9
altitude to \(EM\) in \(\triangle GEM\).\n\n1. **Vertices of the Square**
Rollout 6
$ . [/*]\n<|im_start|>assistant\n<|im_start|>think\nOkay, let me try to work through these
Rollout 14
dx \right) $$ \n<|im_start|>assistant\n<|im_start|>think\nOkay, so I have to prove this inequality
Rollout 5
/100$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. I need to

DOWN_PROJ

Top 16 Positive Activations
Rollout 12
use differentiation under the integral sign, probably Leibniz's rule. \n\nLeibniz's
Rollout 15
Because whether the count is odd or even is all that matters.\n\nWait: exactly:\n\nActually, to get
Rollout 3
focus might specifically align with one temperature.\n\nGiven all that, but as per my knowledge cutoff is 2
Rollout 1
the set S into such's direction.\n\nConsidering all that, after struggling through a number of different avenues,
Rollout 3
: 222 g/mol (3 sig figs)), R is 8.314
Rollout 12
compute f’(ν), perhaps using Leibniz again, since the lower limit is ν,
Rollout 12
Leibniz's rule. \n\nLeibniz's formula for differentiating an integral with variable
Rollout 3
s 169.\n\nTherefore, three sig figs justify 169 m/s.\n\nTherefore
Rollout 3
olar mass is 222 (3 sig figs), R is 8.314
Rollout 1
finite-dimensional space, in order to have equidistant points, they must form a regular simplex, which
Rollout 12
approaching a singularity. So perhaps the Leibniz formula might still apply, but the term involving
Rollout 12
here, even though the first term in Leibniz formula is -F(ν, ν)
Rollout 3
depending on source of problem, maybe requires two sig figs. 222 is three, so
Rollout 3
is 8.314 (4 sig figs), T is 298 K (
Rollout 12
integrand depends on ν.\n\nSo by Leibniz:\n\nf’(ν) = - [
Rollout 3
98). So the final answer with three sig figs: 169 m/s.\n\n Thus
Top 16 Negative Activations
Rollout 5
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 2
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 13
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 0
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 4
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 1
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 6
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 8
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 7
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 3
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 11
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 15
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 9
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 12
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 10
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 14
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user

Layer 6

GATE_PROJ

Top 16 Positive Activations
Top 16 Negative Activations
Rollout 10
in losing positions beyond what I calculated, but Given manual Grundy number calculation up to n=30
Rollout 10
y numbers may be non-periodic.\n\nBut since manual calculation is tedious, let's outline:\n\nGrundy
Rollout 8
sequences: 4P2=12\n\nHappy sequences: All red (needs 2 reds
Rollout 9
$\triangle GEM$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to solve this geometry
Rollout 1
onality condition equality. But then, this seems circular.\n\nAlternatively, setting t = ||x|| squared
Rollout 7
total for original promoted is76*P.\n\nNew promoted (original repeaters with60-6
Rollout 4
this combination of capacitors?\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to find the percentage
Rollout 3
161.4 m/s.\n\nAs an assistant, I need to perhaps suggest both possibilities, but
Rollout 1
equilateral triangle with side length d. Suppose their positions are x1, x2, x3 forming
Rollout 12
\( f'(\nu) \). Using the chain rule and the known derivative of the complete elliptic
Rollout 3
mean (average) speed derived from the Maxwell-Boltzmann distribution:\n\n\[\nv = \sqrt
Rollout 3
.5 m/s.\n\nBut user hasn't specified temperature, but as a standard value, that could vary
Rollout 13
value to the nearest cent.\n<|im_start|>assistant\n<|im_start|>think\nOkay, so Joanie took a $6
Rollout 2
109 mod125. So that matches.\n\nThus, if we follow the steps, the
Rollout 15
be off after some time?\n<|im_start|>assistant\n<|im_start|>think\nAlright, let's try to figure out this
Rollout 15
=3 isn't cyclic.\n\nWait, but earlier thinking process stated there's a cycle, but in reality

UP_PROJ

Top 16 Positive Activations
Rollout 7
$47$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I'm trying to solve this
Rollout 14
dx \right) $$ \n<|im_start|>assistant\n<|im_start|>think\nOkay, so I have to prove this inequality
Rollout 3
n) in m/s\n<|im_start|>assistant\n<|im_start|>think\nOkay, I need to calculate the mean molecular
Rollout 13
value to the nearest cent.\n<|im_start|>assistant\n<|im_start|>think\nOkay, so Joanie took a $6
Rollout 2
1000$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I have this problem here:
Rollout 6
$ . [/*]\n<|im_start|>assistant\n<|im_start|>think\nOkay, let me try to work through these
Rollout 12
[P. Turan]\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to prove that the
Rollout 0
$ be the resulting product?\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I need to figure out how
Rollout 8
$m + n$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. We need to
Rollout 1
mathcal{H}$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to show that there
Rollout 4
this combination of capacitors?\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to find the percentage
Rollout 3
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nCompute the mean molecular speed v in the
Rollout 7
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nIn a mathematics test number of participants is
Rollout 4
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nTwo capacitors with capacitance values $
Rollout 8
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nKathy has $5$ red cards
Rollout 6
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet's say a language $L
Top 16 Negative Activations
Rollout 3
(average) speed derived from the Maxwell-Boltzmann distribution:\n\n\[\nv = \sqrt{\
Rollout 9
0), the altitude to EM would just be the horizontal distance from G to the line EM. Wait,
Rollout 6
for large n, so again, each m corresponds to at most one n. Hence, S_L is
Rollout 0
pair (a,b) with a < b corresponds to the exact opposite pair (b,a) with b
Rollout 14
first moment. There is an inequality called the Heisenberg uncertainty principle which relates position and momentum variances
Rollout 1
if and only if the distance between x1 and x2 is ≤ 2r. Here, distance
Rollout 9
height' in the trapezoid is the horizontal distance between the vertical sides, which is 1
Rollout 0
\). Hence, each coprime pair corresponds to a subset of the prime factors of 20
Rollout 9
is calculated by integrating the difference between the upper and lower lines from \(x = 0\) to \(
Rollout 9
since EM is vertical, the altitude would be the horizontal distance from point G to the line x=1
Rollout 9
10.\n\nWhich is y = (5/h)x + (10 - 50/h).\n\n
Rollout 4
58 nJ. The difference between maximum and nominal is +0.1958 nJ
Rollout 1
d sqrt(2) �� 1.414d >= d, so intersection is non
Rollout 3
mean (average) speed derived from the Maxwell-Boltzmann distribution:\n\n\[\nv = \sqrt
Rollout 9
. The altitude to EM is simply h, the horizontal distance from G to EM.\n\nWait, maybe visual
Rollout 14
, f(x) would be negative, which contradicts the given that $f(x) \geq

DOWN_PROJ

Top 16 Positive Activations
Rollout 0
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 1
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 4
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 7
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 3
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 15
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 12
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 10
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 11
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 8
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 6
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 14
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 9
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 2
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 5
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 13
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Top 16 Negative Activations
Rollout 2
9 \times \cdots \times \underbrace{99\cdots9}_{\text
Rollout 2
9 \times \cdots \times \underbrace{99\cdots9}_{\text
Rollout 12
\nu<1$ . [P. Turan]\n<|im_start|>assistant\n<|im_start|>think\nOkay, so
Rollout 0
, but in reality, since a is a number composed by assigning primes to a (possibly a = product
Rollout 3
's choice. Given a chemistry book, Oxtoby, example problem: At 25°C,
Rollout 14
dx \right) $$ \n<|im_start|>assistant\n<|im_start|>think\nOkay, so I have to prove this inequality
Rollout 7
$47$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I'm trying to solve this
Rollout 6
$ . [/*]\n<|im_start|>assistant\n<|im_start|>think\nOkay, let me try to work through these
Rollout 1
mathcal{H}$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to show that there
Rollout 4
0.88% error, whereas the worst-case scenario would lead to higher.\n\nWait, compute both
Rollout 3
n) in m/s\n<|im_start|>assistant\n<|im_start|>think\nOkay, I need to calculate the mean molecular
Rollout 12
[P. Turan]\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to prove that the
Rollout 0
$ be the resulting product?\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I need to figure out how
Rollout 2
1000$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I have this problem here:
Rollout 1
0.\n\nSo the point y must be equidistant from all points in S, with distance d
Rollout 4
the same direction, which is more of a worst-case scenario. Which does the question expect? The question

Layer 7

GATE_PROJ

Top 16 Positive Activations
Rollout 4
0.88%, or maybe the answer is exact at 0.87%.\n\nBut wait,
Rollout 11
. Hence, their imaginary parts must cancel each other out.\n\nSimilarly, for q, which is a/b
Rollout 3
169: 1.685e2 is 168.5, rounded
Rollout 3
exact)( since formula has 8, which is exact), R=8.314,
Rollout 4
00077 is 7.7e-5. sqrt(7.7e-
Rollout 13
.0044) but that may complicate.\n\nAlternatively, note that (a + b)^
Rollout 14
\] ≤8J K.\n\nThis seems complicate.Playing around expression:\n\nLet’s set variables
Rollout 3
answer expects 170 m/s as an approximate answer. But some precision is required, so
Rollout 3
168.5\n\nThus, answer is approximately 168.5 m/s. Thus
Rollout 3
169.\n\nTherefore, three sig figs justify 169 m/s.\n\nTherefore the valid
Rollout 4
) )^2 )\n\nWait, this is another way of writing the terms. Maybe this will increase precision
Rollout 0
and 144. How is it multiplicative, related to the number of prime factors?\n\nYes
Rollout 3
says "mean".\n\nAlternatively, maybe the problem conflates the two, but the term mean is usually average
Rollout 3
precise calculation using more decimal places.\n\nLet me recalculate the exact value.\n\n8 * 8.3
Rollout 4
Given all the uncertainties are in two significant figures at most, perhaps the answer should be to two significant figures
Rollout 7
.\n\nTherefore, this is impossible. Hence under any circumstances,given these constraintsthe answer to part(b
Top 16 Negative Activations

UP_PROJ

Top 16 Positive Activations
Rollout 1
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\n
Rollout 0
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\n
Rollout 4
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\n
Rollout 12
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\n
Rollout 3
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\n
Rollout 2
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\n
Rollout 8
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\n
Rollout 11
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\n
Rollout 13
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\n
Rollout 15
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\n
Rollout 7
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\n
Rollout 10
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\n
Rollout 9
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\n
Rollout 6
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\n
Rollout 14
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\n
Rollout 5
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\n
Top 16 Negative Activations
Rollout 13
would owe with quarterly compounding.\n\nCalculating the difference: $9,628.24
Rollout 3
28,392.25\n\nDifference between 28,408.1
Rollout 14
the optimal function.\n\nEarlier, in Calculus of Variations, we found that in order to optimize under
Rollout 14
.\n\nFrom calculations before, after applying Calculus of Variations, we obtained the optimal I in this context
Rollout 10
5,7,10,12,...\n\nDifference between 0 and 2: 2\n\n
Rollout 14
here's another thought inspired by the Calculus of Variations result. Earlier when we tried to maximize I
Rollout 8
−r) *(5−r)! (choosing 5−r greens and permuting them
Rollout 1
2:\n\n-⟨x + x', y} { x + x', y} }⟩ +
Rollout 8
10P5. If they are indistinct, then the number of distinct color sequences is
Rollout 8
5 green cards, with each color being indistinct, then the number of distinct color sequences is
Rollout 11
y/z + z/x. So if we can relate q to expressions in terms of x, y,
Rollout 6
? Well, NP is different. In the angel model, you have this poly-length string that can depend
Rollout 14
but maybe I had an error in Calculus of Variations?\n\nWait, perhaps the optimal function for f
Rollout 1
orthonormal system. Thus, if we can identify y as the center of such a system. The
Rollout 12
it states:\n\nThe derivatives of the complete elliptic integrals are:\n\ndK/dk = [ E
Rollout 7
, all scores are increased by5. At this point the average of the promoted participants becomes75 and

DOWN_PROJ

Top 16 Positive Activations
Rollout 1
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 0
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 4
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 7
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 8
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 3
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 9
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 14
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 6
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 10
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 15
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 12
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 11
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 2
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 13
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 5
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Top 16 Negative Activations
Rollout 3
\boxed{169 \text{ m/s}}.\n\nHowever, depending on source of problem,
Rollout 3
three sig figs justify 169 m/s.\n\nTherefore the valid boxed answer is \boxed{
Rollout 3
\(169 \, \text{m/s}\).\n\n**Final Answer:**\n\[\n\
Rollout 3
68.5 \, \text{m/s}\n\]\n\n4. **Rounding to Significant
Rollout 3
0)= approx 161.4 m/s.\n\nAlternatively, precise sqrt(2605
Rollout 3
169} \, \text{m/s}\n\]<|im_end|>
Rollout 3
)= approximately 161.35 m/s.\n\nTherefore, approximately 161.4
Rollout 3
three sig figs is 169 m/s. Thus, present answer boxed as \boxed{
Rollout 3
three sig figs: 169 m/s.\n\n Thus, the mean molecular speed of radon
Rollout 3
answer around \boxed{169} m/s. But need to specify.\n\nAlternatively, here's
Rollout 3
) �� 168.5 m/s\n\nSo approximately 168.5 m
Rollout 3
) is approximately 168.6 m/s. So about 169 m/s.
Rollout 3
Therefore, approximately 161.4 m/s.\n\nSo, depending on the temperature assumed (2
Rollout 3
STP), 161.4 m/s.\n\nAs an assistant, I need to perhaps suggest
Rollout 3
nearest meter per second, 169 m/s. Alternatively, keeping one decimal, 16
Rollout 3
0)= approximately 167.2 m/s. So ~167 m/s.\n\nBut

Layer 8

GATE_PROJ

Top 16 Positive Activations
Rollout 3
n) in m/s\n<|im_start|>assistant\n<|im_start|>think\nOkay, I need to calculate the mean molecular
Rollout 4
this combination of capacitors?\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to find the percentage
Rollout 6
$ . [/*]\n<|im_start|>assistant\n<|im_start|>think\nOkay, let me try to work through these
Rollout 12
[P. Turan]\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to prove that the
Rollout 1
mathcal{H}$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to show that there
Rollout 14
dx \right) $$ \n<|im_start|>assistant\n<|im_start|>think\nOkay, so I have to prove this inequality
Rollout 2
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 6
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 3
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 8
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 7
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 4
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 10
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 11
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 12
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 9
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Top 16 Negative Activations
Rollout 11
a^2 c (since abc = 1 => c = 1/(ab)), so maybe:\n\n
Rollout 8
a random order. She will be happy if and only if all the red cards laid out are adjacent and
Rollout 9
. Since point G is the apex of the triangle opposite the base EM, which is the vertical side of
Rollout 9
0/h <5, because h>10 => 50/h <5. Similarly,
Rollout 5
= 2(b + 75) \implies 3b + 75 =
Rollout 9
So 20y = 100 => y = 5.\n\nTherefore, the y-coordinate
Rollout 5
the legs, has a length equal to the average of the two bases, \( b + 50
Rollout 5
the legs) has a length equal to the average of the two bases. Also, if there is a
Rollout 7
\n\nSo 15P=10N => 3P=2N => P = (
Rollout 5
The segment that joins the midpoints of the legs divides the trapezoid into two regions whose areas
Rollout 0
frac{a}{b}\) where \(0 < a < b\) and \(a\) and \(
Rollout 0
times b = 20!\), \(0 < a < b\), and \(\gcd(a
Rollout 2
9 \equiv 5 \mod 8 \implies 5m + 5 \equiv
Rollout 8
laid out are adjacent. For example, card orders RRGGG, GGGGR, or RRR
Rollout 0
8 assignments:\n\n1. Assign none to a => a=1, b=720 (
Rollout 13
= 1):**\n\[\nA_{\text{annual}} = 6000 \

UP_PROJ

Top 16 Positive Activations
Rollout 3
n) in m/s\n<|im_start|>assistant\n<|im_start|>think\nOkay, I need to calculate the mean molecular
Rollout 1
mathcal{H}$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to show that there
Rollout 12
[P. Turan]\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to prove that the
Rollout 4
this combination of capacitors?\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to find the percentage
Rollout 14
dx \right) $$ \n<|im_start|>assistant\n<|im_start|>think\nOkay, so I have to prove this inequality
Rollout 11
.\n\n*David Altizio*\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I have this problem here where
Rollout 6
$ . [/*]\n<|im_start|>assistant\n<|im_start|>think\nOkay, let me try to work through these
Rollout 9
$\triangle GEM$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to solve this geometry
Rollout 8
$m + n$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. We need to
Rollout 7
$47$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I'm trying to solve this
Rollout 13
value to the nearest cent.\n<|im_start|>assistant\n<|im_start|>think\nOkay, so Joanie took a $6
Rollout 2
1000$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I have this problem here:
Rollout 0
$ be the resulting product?\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I need to figure out how
Rollout 15
be off after some time?\n<|im_start|>assistant\n<|im_start|>think\nAlright, let's try to figure out this
Rollout 5
/100$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. I need to
Rollout 10
matter how his opponent plays.\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to prove that in
Top 16 Negative Activations
Rollout 13
18242*1.03≈≈1.3047731
Rollout 13
72322*1.03≈1.51258972
Rollout 13
41492*1.03≈1.55796741
Rollout 8
240=31/126≈0.246.\n\nThus, the answer
Rollout 13
37789*1.03≈1.34391637
Rollout 13
88531*1.03≈1.42576088
Rollout 13
86923*1.03≈1.38423386
Rollout 0
a,b) where a < b. So the count is half the number of coprime ordered pairs
Rollout 3
6/0.697433≈27,950. sqrt(2
Rollout 10
(Grundy0); mex{0}=1→Grundy[21}=1\n\nn=
Rollout 13
77008*1.03≈1.26677008
Rollout 10
(Grundy0); mex{0}=1→Grundy[16}=1\n\nn=
Rollout 13
3800310242≈1.30477318
Rollout 13
71188*1.03≈1.46853371
Rollout 10
(Grundy0); mex{0}=1→Grundy[18}=1\n\nn=
Rollout 13
0881*1.03 �� 1.1255088

DOWN_PROJ

Top 16 Positive Activations
Rollout 15
", which is the same as parity (sum neigbors mod2 =1). Therefore, the
Rollout 2
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 13
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 0
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 6
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 1
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 7
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 9
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 4
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 3
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 15
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 8
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 12
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 14
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 11
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 10
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Top 16 Negative Activations
Rollout 5
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nOne base of a trapezoid
Rollout 0
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nGiven a rational number, write it as
Rollout 2
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nFind the remainder when $9 \times
Rollout 1
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet $ \mathcal{H
Rollout 4
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nTwo capacitors with capacitance values $
Rollout 7
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nIn a mathematics test number of participants is
Rollout 8
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nKathy has $5$ red cards
Rollout 3
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nCompute the mean molecular speed v in the
Rollout 13
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nJoanie takes a $\$6,\
Rollout 6
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet's say a language $L
Rollout 10
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nConsider the following two person game. A
Rollout 11
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nSuppose $a,\,b
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<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nSquare $AIME$ has sides of
Rollout 14
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nA function $f:[0,\
Rollout 12
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nProve that the function \[ f
Rollout 15
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet $n\geq1

Layer 9

GATE_PROJ

Top 16 Positive Activations
Rollout 7
equations, but led to impossibility. Thus, perhaps the answer is different.\n\nAlternatively, maybe N=
Rollout 8
.\n\nSame for k=3.\n\nWait, so perhaps for each k from 1 to 4,
Rollout 6
. But this is impossible in polynomial time.\n\nSo perhaps the answer is not straightforward, hence why reference to
Rollout 9
has a portion inside and a portion outside. So perhaps when h is smaller than 10? Wait
Rollout 1
their radius is d / sqrt(2}. So perhaps the pairwise distance between sphere centers squared is much larger
Rollout 14
. Thus the inequality is actually not tight. So perhaps the 8 is not optimal. But the problem
Rollout 11
a polynomial must have non-real coefficients. Hence, perhaps looking for cases where p and q have specific relationships
Rollout 15
, then those n are the solutions. So, perhaps powers of two minus one? Because n=1
Rollout 15
is insufficiently modeling the actual system. Hence, perhaps equivalence is not correct.\n\nBut why this discrepancy?
Rollout 3
I can't compute an exact value. Therefore, perhaps the problem is in a context where they expect the
Rollout 6
p(n), which is not feasible.\n\nSo, perhaps the encoding has to be different. Let's think
Rollout 7
there are constraints I haven't considered.\n\nWait, perhaps non-integer averages? But average A=9
Rollout 14
, perhaps using Hölder's for two functions. Think of the three-factor inequality.\n\nAlternatively if we consider
Rollout 11
is symmetric in a, b, c. So perhaps we can relate p and q through symmetric expressions.\n\n
Rollout 11
that still real c. No good.\n\nTherefore, perhaps the only way is to not have a and b
Rollout 10
the game each move is subtract a square, so perhaps using octal games concept or mathematical induction.\n\nAlternatively
Top 16 Negative Activations
Rollout 1
Hilbert space, let $ d>0$ , and suppose that $ S$ is
Rollout 1
distance between any two distinct points in $ S$ is equal to $ d$ . Show
Rollout 1
>0$ , and suppose that $ S$ is a set of points (not necessarily count
Rollout 1
$ S$ is equal to $ d$ . Show that there is a point $ y
Rollout 1
user\nLet $ \mathcal{H}$ be an infinite-dimensional Hilbert space, let
Rollout 1
able) in $ \mathcal{H}$ such that the distance between any two distinct points
Rollout 1
$ y\in\mathcal{H}$ such that \n\[ \left\{\frac
Rollout 0
floor(20/49)=2 +0=2\n\nPrimes 11: floor
Rollout 1
points (not necessarily countable) in $ \mathcal{H}$ such that the distance
Rollout 12
(1 -k2)K(k) ] / (k(1 -k2))\n\nTherefore,
Rollout 0
). For example, primes between N/2 and N cannot come in pairs. So hence exponents for
Rollout 0
floor(20/25)=4 +0=4\n\nPrime 7: floor(2
Rollout 12
interval $ 0<\nu<1$ . [P. Turan]\n<|im_start|>assistant\n
Rollout 2
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nFind the remainder when $9 \times
Rollout 4
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nTwo capacitors with capacitance values $
Rollout 3
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nCompute the mean molecular speed v in the

UP_PROJ

Top 16 Positive Activations
Rollout 1
: Let me define y such that y = sum_{i=1}^n a_i x_i
Rollout 14
�� A ^infty f(x) dx <= ��A^infty (x f(x
Rollout 14
the convexity inequality (a +b)^3 <=4 (a3+ b3), but need
Rollout 14
A.\n\n then cube we have:\n\nI^3 <= (sqrt( J) sqrt(A ) + K
Rollout 1
need to find y such that y ∈ ��_{x ∈ S} sphere(x, d/sqrt
Rollout 8
card sequences = 2 *5! + Sum_{r=1}^4 [2 * P
Rollout 14
A f(x).\n\nHence, f(x) <= (x f(x))/A.\n\nTherefore, I
Rollout 8
total happy sequences = 240 + Sum_{k=1}^4 [2 * (
Rollout 15
=2 when starts from OO it cycles. But over GF(2), for no.cycles:\n\nBut
Rollout 14
dx = �� f(x) *1 dx <= (�� (f(x))^2 x^a
Rollout 14
/ sqrt(J).\n\nPlug this into upper bound I <= sqrt(J) sqrt(A ) + K/A.\n\n
Rollout 14
} \cdot x^{-1/3}| dx ≤ ||f||_3 ||x^{1/
Rollout 14
2} *x^{-1/2} dx <= ||f(x)x^{1/2} ||
Rollout 1
}{(x - y) ||·|| (sqrt(2}/d}{(x' - y
Rollout 0
the set of prime factors, then a = product_{p in S} p^{e_p}, and
Rollout 14
>=A in this region, x f(x) >=A f(x).\n\nHence, f(x)
Top 16 Negative Activations
Rollout 6
) to decide \( x \in L \).<|im_end|>
Rollout 0
Final Answer**\n\boxed{128}\n\n<|im_start|>answer\nGiven a rational number between 0 and
Rollout 2
is \(\boxed{109}\).<|im_end|>
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**\n\boxed{(-1, 3)}\n\n<|im_start|>answer\nGiven three complex numbers \(a, b
Rollout 5
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nOne base of a trapezoid
Rollout 0
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nGiven a rational number, write it as
Rollout 2
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nFind the remainder when $9 \times
Rollout 1
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet $ \mathcal{H
Rollout 3
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nCompute the mean molecular speed v in the
Rollout 4
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nTwo capacitors with capacitance values $
Rollout 8
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nKathy has $5$ red cards
Rollout 10
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nConsider the following two person game. A
Rollout 6
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet's say a language $L
Rollout 9
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nSquare $AIME$ has sides of
Rollout 7
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nIn a mathematics test number of participants is
Rollout 11
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nSuppose $a,\,b

DOWN_PROJ

Top 16 Positive Activations
Rollout 10
the pattern is adding 2 then 3 alternately. Hmmm. If this continues, then the
Rollout 10
**Understanding the Game**:\n - Players take turns removing square numbers (1, 4,
Rollout 11
zero, but in a non symmetric way.\n\nAlternately, consider all three variables with the same modulus,
Rollout 1
istant set have to be of this form?\n\nAlternately, let me try to create y. Assume that
Rollout 10
opponent plays, we analyze the game where players alternately remove a square number of pebbles. The
Rollout 10
on the table. Two players make their moves alternately. A move consists of taking off the table
Rollout 10
me try to understand the game first.\n\nPlayers take turns removing a square number of pebbles (1
Rollout 10
Hmm.\n\nAlternatively, looking at their parity: alternately even and odd. But maybe every other pair.
Rollout 14
1=∫0^AA f2 dx.\n\nSimilarly, for I2= ��A^in
Rollout 10
's a regular pattern of losing positions generated by alternately adding 2 and 3, resulting in an
Rollout 14
applications?\n\nWait, let's see. Let's denote:\n\n$I = \int_0^\infty
Rollout 13
multiplying by 1.1255. Alternatively, perhaps use the standard multiplication grid. But this
Rollout 13
then adjust. Wait, that's too big. Alternatively, use written multiplication.\n\nLine up 1.
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0.\n\nn -s≡0 or1.\n\nSimilarly, maybe analysis similar to NIM in mod4
Rollout 1
2 - ||x2||2)/2.\n\nSimilarly, using the distance between x1 and x2
Rollout 12
, use a trigonometric substitution.\n\nLet’s denote t = ν coshθ. Wait, but
Top 16 Negative Activations
Rollout 7
0). Their new scores are original+5 (<=64). Average should be47. So
Rollout 14
for each x, we have t f(x) <= (f(x)^2)/2 + t^
Rollout 12
coshθ would be problematic as coshθ >=1. Instead, hyperbolic functions may not
Rollout 14
1 and K=1, then I^3 <=8. The earlier optimization using a linear function with
Rollout 14
) dx.\n\nHölder tells us: I <= (��f^p )^{1/p}
Rollout 3
K (unless stated otherwise), perhaps answer around \boxed{169} m/s. But need
Rollout 14
for any a,b>0,\n\nI^3 <= [(4 a^{-1}) J + (2
Rollout 5
the correct answer. \n\n**Final Answer**\n\boxed{181}\n\n<|im_start|>answer\nLet the
Rollout 8
, 5! card sequences.\n\nIf 1<=r<=4:\n\nThen, the color sequences must
Rollout 14
.\n\nTerm2: a/2.\n\nSo I <= sqrt( 2 K a ) + a/
Rollout 8
- For each possible number of red cards (0 <= r <=5):\n\nIf r=0: all
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with original scores <60.\n\nTherefore, x <=N/3, and (N/3 -
Rollout 14
, use integrating in terms of K:\n\nSince x >=A in this region, x f(x) >=
Rollout 13
7.12.\n\n**Final Answer**\n\boxed{187.12}\n\n<|im_start|>answer
Rollout 0
answer is 128, corresponding to \boxed{128}? Or wait, wait
Rollout 8
all green or all red).\n\nBut for 1 <= r <=4: 2 * P(5

Layer 10

GATE_PROJ

Top 16 Positive Activations
Rollout 12
of power functions. But not sure. Let me recall the Beta function:\n\nB(p, q) =
Rollout 15
ent over GF(2).\n\nNow, let me recall: for the adjacency matrix of a path graph,
Rollout 15
Alternatively, test specific values.\n\nWait, let's recall initially tested n=1 good, n=2
Rollout 12
differentiating in a certain way.\n\nAlternatively, I recall that in differentiation under the integral sign, if the
Rollout 3
to be careful here. Let me check.\n\nI recall the root mean square (rms) speed formula is
Rollout 5
/100. \n\nFirst, let me recall some trapezoid properties. In a tr
Rollout 14
a weighted Hölder's inequality. \n\nLet me recall: in a weighted Hölder's inequality, for
Rollout 12
integrals, the derivative is known. Let me recall that the complete elliptic integral of the first kind
Rollout 1
to use the parallelogram law. Let me recall that in a Hilbert space, given two vectors
Rollout 1
are mutually orthogonal after scaling.\n\nWait, let me recall that if you have a regular simplex in R^n
Rollout 5
quickly. \n\nAlternatively, perhaps it is quicker to recall the formula for the line that divides the trape
Rollout 12
term doesn't actually contribute?\n\nWait, let's recall that when we use differentiation under the integral sign,
Rollout 5
which is sort of a weighted average. Let me recall that if you have a line parallel to the bases
Rollout 1
is zero for distinct x, x'.\n\nLet me recall that in Hilbert spaces, if we have a
Rollout 7
(P +x)\n\nFrom previous steps:\n\nLet's write this out:\n\n76P + (A +
Rollout 14
with exponents that are conjugate. Let me recall Hölder's inequality: for $p$ and
Top 16 Negative Activations
Rollout 1
= x0 + z' is ||z - z'|| = d. Now, we need to find
Rollout 1
H}$ such that the distance between any two distinct points in $ S$ is equal to
Rollout 1
S' is ||z|| = ||x - x0||, but in the original set S,
Rollout 1
x'||2 = 0 ⇒ ⟨x - x', t⟩ = ||x||2 - ||
Rollout 1
, x' ∈ S, then ⟨x - x', y⟩ = ⟨x, y⟩ -
Rollout 1
sqrt(2),\n\nwe also have ||x - x'|| = d = sqrt(2) * ||
Rollout 1
inner product formula:\n\nd2 = ⟨x - x', x - x'⟩ = ||x||
Rollout 1
d / sqrt(2) ≥ ||x - x'||, which is d sqrt(2) ≥
Rollout 1
2.\n\nWhich gives that inner product ⟨x - x', y⟩ = (||x||2 -
Rollout 1
for some constant c. Then, ⟨x - x', y⟩ would be (⟨x, x
Rollout 1
:\n\nFor each z ∈ { z = x - x0 : x ∈ S }, we have ||z
Rollout 1
x0 ∈ S. Let z = x - x0 for x ∈ S. Then, S =
Rollout 1
then from the above equation, Re⟨x - x', y⟩ = 0.\n\nBut in general
Rollout 1
||x||2 = c and ||x - x'|| = d).\n\nAlso, ⟨x, y
Rollout 1
and x' in S. Then ||x - x'|| = d. Let me assume that there exists
Rollout 1
x'||2 = d2 ⇒ ⟨x - x', x - x'⟩ = d2 ⇒

UP_PROJ

Top 16 Positive Activations
Rollout 11
π. Forcing abc=1.\n\nHowever, suppose arranging a =k e^{iθ}, b
Rollout 1
)/2.\n\nPutting all this together.\n\nNow, suppose that we can model some kind of system.\n\nLet
Rollout 14
the function such that multiple variables are involved.\n\nSuppose I make a substitution x= cy where c is
Rollout 14
) such that the problem constraints are respected.\n\nSuppose a choose w(x)= x. Then lhs\n\n
Rollout 14
yet again the first integral diverges. However, suppose we split the 1/x integral between \ the
Rollout 1
+ ⟨y, y⟩ = 0.\n\nWhich is equivalent to:\n\n⟨x, x'⟩
Rollout 14
bound I^3 ≤8 J K.\n\nSuppose we claim that for any a,b>0,\n\n
Rollout 14
.\n\nAnother approach inspired by weighted integrals:\n\nSuppose we set up a weight function that balances between
Rollout 14
. But to involve the given integrals.\n\nSuppose set a=3 and b=3/2
Rollout 11
1, 3). Maybe try this.\n\nSuppose let’s think when p = -1 and q
Rollout 11
not have a and b conjugates. Alternatively, suppose three complex numbers with arguments ��, ��
Rollout 1
y||2 = d2 / 2.\n\nWhich can be written as:\n\n-2⟨x,
Rollout 1
/ 2 + d2/ 2.\n\nWhich is equal to:\n\n( d2/2 -
Rollout 14
the weight a term involving x. For example, suppose I take the weight to be (1 + x
Rollout 11
terms. It may have some symmetry properties.\n\nSuppose t1 and t2 satisfy t2 =1
Rollout 11
equal to the conjugate of another term:\n\nSuppose a/b = \overline{(b/c)},
Top 16 Negative Activations
Rollout 14
0,\infty)$ is integrable and $$ \int_0^\infty f(x
Rollout 11
$c$ are three complex numbers with product $1$ . Assume that none of $
Rollout 15
lamps are placed in a line. At minute 0, some lamps are on (maybe all of
Rollout 5
zoid into two regions with areas in the ratio 2:3. Then, we need to find
Rollout 11
$c$ are real or have absolute value $1$ . Define\n\begin{tab
Rollout 9
there's a square named AIME with sides of 10 units. There's also an isos
Rollout 10
table $x$ pebbles where $x$ is the square of any positive
Rollout 11
where none of them are real or have absolute value 1, we need to find the possible values of
Rollout 9
-right. So the square is AIME with sides 10 units each. Then we have an is
Rollout 1
two things: 1) each vector has norm 1, and 2) any two distinct vectors
Rollout 11
numbers \(a, b, c\) with product 1, where none of them are real or have
Rollout 11
pose $a,\,b,$ and $c$ are three complex numbers with product
Rollout 1
thonormal system. That means each vector has length 1 and any two distinct vectors are orthogonal.\n\nFirst
Rollout 1
}(x - y) \) must have norm 1. This implies \( \|x - y\
Rollout 12
taken) is monotonically decreasing in the interval $ 0<\nu<1$ .
Rollout 9
. Since it's a square, all sides are 10 units. Let's label the square with

DOWN_PROJ

Top 16 Positive Activations
Rollout 11
\overline{z} is real (the modulus squared). Since p is a combination of a,
Rollout 11
So they can’t be real, even if their modulus is not 1. So all of a,
Rollout 11
c= e^{-i2θ}. But modulus of a and b is 1, which is prohibited
Rollout 1
the distance between any two is the same, they form a regular simplex. The regular simplex has the property
Rollout 11
1. Let k be a complex number with modulus not 1, e.g., k =
Rollout 1
But actually no, in the previous, if we take multiple points, we need to ensure that they lie
Rollout 11
has modulus 1 and argument 0, so modulus r_a r_b r_c = 1,
Rollout 1
Hilbert space \( \mathcal{H} \) and a set \( S \) of points
Rollout 15
a specific case for particular structures.\n\nIn particular, considering the path graph's adjacency matrix is tridiagonal
Rollout 1
non-empty.\n\nSo in our case, if we take any finite subset of S, then the intersection of
Rollout 11
b, c are complex but non-real, with modulus different from 1, but their product is
Rollout 1
d2 / 2 = d2. Then considering squared distances in the planes, the intersection would be
Rollout 3
answer\nTo compute the mean molecular speed \( v \) of radon (Rn) gas,
Rollout 1
a constant.\n\nBut in that case, if we take x, x' ∈ S, then ⟨x
Rollout 1
certain conditions are satisfied.\n\nWait, for example, take three points in R^3 with equal pairwise distances
Rollout 1
can be considered as an orthonormal basis for some subspace of H. However, since H is
Top 16 Negative Activations
Rollout 12
is:\n\nF(φ, k) = ��_0^φ [ 1 / sqrt(
Rollout 9
_upper and y_lower from a to b is ��[a to b] (y_upper - y
Rollout 12
is:\n\nF(φ, k) = ��_0^phi [ 1 / sqrt(
Rollout 12
to1. So f(ν) = ��_ν^1 [1 / sqrt( (
Rollout 12
). Start with:\n\nf(ν) = ��_{ν}^1 [ (t2 -
Rollout 12
= ν] * dν/dν + ��_{ν}^1 ��/∂ν
Rollout 9
+50/h ) ] dx\n\n= ��[0 to10] [5/h x
Rollout 12
Let me write:\n\nf(ν) = ��_{ν}^1 [ (t2 -
Rollout 12
)^{-1/2} ) ] + ��_{ν}^1 ( ��/∂
Rollout 12
is written as:\n\nf(ν) = ��_{ν}^1 \frac{1}{
Rollout 12
/ (1 - k2 sin2 theta)^{3/2} } ] d theta\n\nBut
Rollout 12
Wait let me verify:\n\nK(k) = ��0^{pi/2} [1 / sqrt
Rollout 12
1. Therefore,\n\nf(ν) = ��_{ν2}^1 [ (s -
Rollout 3
16 * 0.222 �� 0.697. So 1
Rollout 12
the problem is now f(ν) = ��_{ν}^1 1 / sqrt(
Rollout 12
. Then,\n\nIntegral f(nu) = ��_{0}^{π/2} [ d

Layer 11

GATE_PROJ

Top 16 Positive Activations
Rollout 8
want to cross-verify.\n\nWait, let’s think of another approach. For instance, imagine that we
Rollout 9
to EM, h can be anything. Let's think. If h is less than 10,
Rollout 1
be uncountable. Hmm.\n\nBut let's think: Suppose that S is a set where all pairwise
Rollout 15
check if n=4 is good. Let me think of an initial state. For example, starting with
Rollout 10
at n=25. However, let's think modulo4.\n\nSquares mod4 are 0
Rollout 15
n=1=2^0. Let's think of our n= 2^k.\n\nFor
Rollout 15
model the transitions over GF(2). Let me think for n=3.\n\nThe update rule over GF
Rollout 2
25.\n\nThis seems more challenging. Let me think.\n\nWe need to compute P mod125
Rollout 12
substitution φ where t = cosθ? Let me think. For if t is from ν to 1
Rollout 10
odd. But maybe every other pair. Let me think. Alternatively, maybe in each two steps, the
Rollout 11
Perhaps we can use some relationships here. Let's think about using the fact that if z is a non
Rollout 1
the same Hilbert space.\n\nWait, let me think again.\n\nGiven set S wherein between any two points
Rollout 10
Wait, here's an alternative idea. Let me think about parity:\n\nSuppose we have n as an
Rollout 11
3). Maybe try this.\n\nSuppose let’s think when p = -1 and q=3.\n\n
Rollout 15
linear algebra problem over GF(2). Let me think.\n\nGiven that the next state is determined by whether
Rollout 3
which one is the actual mean speed. Let me think.\n\nMean speed, or average speed, is indeed
Top 16 Negative Activations
Rollout 14
a )(a/2 ) +3 (sqrt(2 K a ))(a2/4 )
Rollout 14
cube we have:\n\nI^3 <= (sqrt( J) sqrt(A ) + K/A )^
Rollout 14
+ a/2.\n\nNeed to show ( sqrt(2 K a ) +a /2 )3
Rollout 1
need to find a point y such that { sqrt(2)/d (x - y ) }x
Rollout 1
e_i} where each e_i is (sqrt(2)/d)(x_i - y) for
Rollout 14
} J^{1/4} + (sqrt( J ) )/2.\n\nBut we are to
Rollout 14
2: a/2.\n\nSo I <= sqrt( 2 K a ) + a/2.\n\n
Rollout 3
3≈26,050\n\nsqrt(26,050)= approx 1
Rollout 3
3≈27,950\n\nsqrt(27,950)= approximately 1
Rollout 14
: K/A= K/(2K /sqrt( J )) )=sqrt( J )= /
Rollout 3
,408.123\n\nsqrt(28,408.123
Rollout 14
/2.\n\nExpand S3:\n\n= [sqrt(2 K a ) ]3 +3 [sqrt
Rollout 3
6,034.21.\n\nsqrt(26,034.21)=
Rollout 4
(from the voltage), gives for energy:\n\nsqrt( (0.0036055
Rollout 1
a point y such that when I take (sqrt(2)/d)(x - y), these vectors
Rollout 12
established that f(nu) = K( sqrt(1 - nu2) )\n\nNow, the derivative

UP_PROJ

Top 16 Positive Activations
Rollout 14
/ k^3. So it's even then not tight. Therefore, maybe the 8 here
Rollout 11
Even this direction is not leading anywhere. Finally, think of the answer. Since in olympiad problems
Rollout 2
*99*999*... So thinking about modulus 1000, when you
Rollout 9
{25} . Seems done.\n\nThat's logical. Another way:\n\nFrom h=25,
Rollout 10
show both even and odd losing positions.\n\nAlternatively, think in terms of how many winning moves leads from each
Rollout 8
are adjacent and all greens are adjacent.\n\nAlternatively, think of the problem as the laid-out 5 cards
Rollout 9
.e., to the left of the square. So then, the triangle connects (negative,5), (
Rollout 6
hash of α_n. Not sure.\n\nWait, think about P_angel: the original TM M is
Rollout 11
+ ca) + 2. Wait, so then p = (1 + a)(1 + b
Rollout 14
4 / k^3. So it's even then not tight. Therefore, maybe the 8
Rollout 15
n \) that are powers of two.\n\nCheck with n=1 (power of two: 2
Rollout 15
utner says a wolfram cite?\n\nOk i think I need to take this carefully: maybe the rule
Rollout 1
The problem is over the Hilbert space, so thinking as in R^n, for which we know that
Rollout 15
\( k \).\n\n**But Wait:**\nAfter reconsidering from some papers and logical reasoning, in linear
Rollout 0
have a < b or a > b, so then to get the (a,b) with a <
Rollout 11
, making it hard to apply that.\n\nAlternatively, think of log variables? If abc =1, set
Top 16 Negative Activations
Rollout 14
We need to show that $I^3 \leq 8 J K$.\n\nHmm. So
Rollout 14
me see.\n\nWe need to relate I^3 ≤8 J K.\n\nBy AM_ GM,\n\nI
Rollout 14
1. Then need to show that I^3 ≤8 *1 *1→ I ≤2.\n\n
Rollout 14
=t. Then the inequality states that I^3 ≤8 *1 *t.\n\nTo make this work
Rollout 14
We need to show that \( I^3 \leq 8JK \).\n\nWe split the integral
Rollout 14
but the question states to prove that I^3 ≤8 J K. Here, if I≈1
Rollout 14
1. But then the inequality requires I^3 ≤8J K, which would give 8 as
Rollout 14
inequality $(a + b + c)^3 \leq ...$. Hmm, but not directly applicable here
Rollout 14
then our goal reduces to showing that I^3 ≤8, which is equivalent to I≤2.
Rollout 14
fty f(x) dx \right)^3 \leq 8\left(\int_0^\
Rollout 14
ity to show:\n\n\[\nI^3 \leq 8JK.\n\]\n\nThus, the
Rollout 9
triangle might be entirely inside the square (if h ≤10) but h is positive.\n\nBut wait
Rollout 14
that we have to find an bound I^3 ≤8 J K.\n\nSuppose we claim that for
Rollout 14
3, so leading to the inequality I^3 ≤8 JK with equality when LHS=1 <
Rollout 14
fty f(x) dx \right)^3 \leq 8\left(\int_0^\
Rollout 14
using the fact that (��f dx)^2 ≤ (��1^2 dx)(∫f^

DOWN_PROJ

Top 16 Positive Activations
Rollout 7
16x =71N -209N/3= (213N -
Rollout 7
x.\n\nTherefore, the equation becomes:\n\n209N/3 +16x =71
Rollout 7
3= (213N -209N)/3=4N/3.\n\nTherefore
Rollout 5
formula:\n\ny = [150 ± sqrt(22500 + 500
Rollout 0
, a can vary from 1 up to sqrt(20!) if possible, but since 2
Rollout 1
a regular simplex in R^n with edge length sqrt(2), then the vectors from the centroid to the
Rollout 7
71N.\n\nNow, subtract 209N/3 from both sides:\n\n16x
Rollout 1
between any two points in the simplex would be sqrt(2) times the radius.\n\nIndeed, if you
Rollout 3
3≈26,050\n\nsqrt(26,050)= approx 1
Rollout 1
a point y such that when I take (sqrt(2)/d)(x - y), these vectors
Rollout 3
,408.123\n\nsqrt(28,408.123
Rollout 1
d2, and their radius is d / sqrt(2}. So perhaps the pairwise distance between sphere centers
Rollout 3
6,034.21.\n\nsqrt(26,034.21)=
Rollout 1
y so that ||y|| = d / sqrt(2), then these vectors would not be orthogonal.\n\n
Rollout 1
our case, each sphere has radius d / sqrt(2} approx 0.707d
Rollout 7
3 +59N/3=209N/3.\n\nSimilarly, the x terms:\n\n
Top 16 Negative Activations
Rollout 4
\pm 0.02 \mathrm{~V}$. What is the percentage error in the
Rollout 4
1200 = let's compute that more accurately.\n\n4.327 / 12
Rollout 11
) sin2θ = 0\n\nExpress sin2θ as 2 sintheta cos theta:\n\n3
Rollout 12
0^{pi/2} [ (k sin2 theta ) / (1 - k2 sin2
Rollout 5
= 181.25, greatest integer not exceeding 181.25 is
Rollout 4
\approx 0.8775\%\n\]\n\nRounding to two significant figures, the
Rollout 12
ν2 + (1 - ν2) sin2θ. Then when t=ν, sinθ
Rollout 11
sinθ + (15/4) sin2θ = 0\n\nExpress sin2θ as
Rollout 12
nu2 + (1 - nu2) sin2θ = 1 - (1 - nu2
Rollout 12
ν2 + (1 - ν2) sin2θ). Differentiating both sides with respect to �
Rollout 12
- (1 - nu2)(1 - sin2θ ) = 1 - k2 cos2
Rollout 12
1 / sqrt(1 - k2 sin2θ) ] dθ\n\nBut alternatively, there
Rollout 12
nu2 + (1 - nu2) sin2θ ) ] dθ\n\nHmm. Let's
Rollout 12
1 / sqrt(1 - k2 sin2θ) ] dθ\n\nOr in another form
Rollout 12
2 theta ) / (1 - k2 sin2 theta)^{3/2} } ] d
Rollout 3
is approximately 222 (since the most stable isotope is radon-222).

Layer 12

GATE_PROJ

Top 16 Positive Activations
Rollout 3
which is 300 - 2, so subtract 66.512 *
Rollout 15
.\n\nL1: neighbor L2 (on). So exactly one neighbor on. Therefore, turns on.\n\n
Rollout 12
since k= sqrt(1 -nu2), so 0 <nu <1 implies k in (
Rollout 10
subtract 9: leaves 0 (losing). So n=9 is winning.\n\nn=10
Rollout 10
: 1, leaving 5 (losing). So n=6 is winning.\n\nn=7:
Rollout 15
ighbors: only L2, which is on. So exactly one neighbor on. So L3's next
Rollout 10
win), subtract 9=6 (win). So all moves lead to opponent's win. Thus n
Rollout 15
L3 (on). Exactly one neighbor on. So L2's next state is on.\n\nFor L
Rollout 10
);n -4=20 (L). So W.\n\nn=25: n -2
Rollout 4
5,025,000, so 5,025,000
Rollout 10
take 4, leave 0 (losing). So n=4 is winning.\n\nn=5:
Rollout 10
W); n -9=3 (W). So L.\n\nn=13: n -1
Rollout 15
:\n\nL1: neighbor L2 (off). So off.\n\nL2: neighbors L1 (off
Rollout 15
:\n\nL1: neighbor L2 (off). So turn off.\n\nL2: neighbors L1 (
Rollout 10
W); n -9=1 (W). So L.\n\nn=11: n -1
Rollout 10
W); n -4=3 (W). So L.\n\nn=8: n -1=
Top 16 Negative Activations
Rollout 4
's start by recalling how to calculate the equivalent capacitance of capacitors in series and then the energy stored
Rollout 14
warz inequality with appropriately chosen vectors / weighet\n\n**Final Answer**\n\boxed{ \left(\int
Rollout 1
as described}}" or as per problem statement.\n\n**[Final Answer]**\n\boxed{\text{Such
Rollout 7
average 47.\n\nLet me try to work through that similarly.\n\nFirst, the original setup: All
Rollout 3
K (unless stated otherwise), perhaps answer around \boxed{169} m/s. But need
Rollout 14
function of x. In the integrals:\n\n- [I] = ��f dx ~ [f
Rollout 3
70 m/s, considering rounding, but original exact formula gives 168.5.\n\nBut
Rollout 4
that as well.\n\nThe formula for the equivalent capacitance in series is C_eq = (C1 *
Rollout 9
it's shaped like a quadrilateral?\n\nLet me sketch mentally: triangle GEM has vertex G outside on
Rollout 7
with part (a). Let's split the problem step by step.\n\nFirst, let's consider the original
Rollout 9
, perhaps my assumption is incorrect. Let's recast this. I need to check whether when h increases
Rollout 11
, q is not real, which violates the problem conditions. Hence, such a choice of a=b,
Rollout 3
that for 298 K.\n\nV_rms = sqrt(3RT/M) = sqrt(
Rollout 4
let's start by recalling how to calculate the equivalent capacitance of capacitors in series and then the energy
Rollout 4
, for capacitors in series, the equivalent capacitance C_eq is given by 1/C_eq =
Rollout 4
7%.\n\nBut wait, let's go step by step for verification once more.\n\nFirst, Compute C_eq

UP_PROJ

Top 16 Positive Activations
Rollout 14
of integrals and use some sort of dual Hölder applications?\n\nWait, let's see. Let's
Rollout 14
apply some weight.\n\nAlternatively consider using a weighted Hölder's inequality. \n\nLet me recall: in a
Rollout 14
. Thus inequality gives I <= �� * �� which is useless.\n\nAlternatively, to solve the problem
Rollout 14
. \n\nLet me recall: in a weighted Hölder's inequality, for positive weight function w:\n\n\
Rollout 10
concept or mathematical induction.\n\nAlternatively, is there an upper limit on the number of moves? Given that each
Rollout 14
J K$. So, it's perhaps getting an upper bound on $I^3$ in terms of
Rollout 3
average speed. But given that different sources can have different conventions, perhaps use both.\n\nBut no, as
Rollout 14
J K, which would give 8 as the upper bound. Because when J=1, K=
Rollout 14
. This suggests that perhaps the optimal function that gives equality is a different one. Maybe a function with compact
Rollout 1
the previous calculation into the orthogonality condition equality. But then, this seems circular.\n\nAlternatively,
Rollout 1
d>0$ , and suppose that $ S$ is a set of points (not necessarily
Rollout 1
answer\nGiven an infinite-dimensional Hilbert space \( \mathcal{H} \) and a set
Rollout 0
Therefore, odd. Hence sqrt(N!) is not integer. Hence N! is not a square. Therefore
Rollout 14
1||infty, but i think triple Hölder not in that manner.\n\nAnother approach. Think variables
Rollout 2
by (-1), etc.\n\nWait, but the third term is999≡-1 mod1
Rollout 14
1/r=1. Maybe apply three-term Hölder.\n\nWriting (�� f(x) dx )^
Top 16 Negative Activations
Rollout 3
as earlier for radon)\n\nDenominator: pi*0.028= ~0.0
Rollout 14
tells us: I <= (��f^p )^{1/p} ( ��1^q
Rollout 3
.15 K):\n\nCompute as earlier:\n\nsqrt(8*8.314*27
Rollout 3
Rn at 298 K is sqrt(3RT/M). Let's compute:\n\n3*
Rollout 12
ν) / ( (1/ν) sqrt( (t2 - ν2)(1 - t
Rollout 12
x = 1/ν of dx over sqrt( (x2 -1)(1 - ν2
Rollout 12
0. The integrand becomes:\n\n1 / sqrt(1 - k2 cos2theta) =
Rollout 1
our case, each sphere has radius d / sqrt(2} approx 0.707d
Rollout 3
3≈27,950. sqrt(27,950)=167
Rollout 12
is ��0^phi 1 / sqrt(1 - k2 sin2theta) dtheta
Rollout 12
, and the integrand becomes:\n\n1 / sqrt( ( (ν + u)^2 - ν2
Rollout 3
3≈28,622\n\nsqrt(28,622)= about 1
Rollout 3
m/s. Let's check my math:\n\nsqrt(8*8.314*29
Rollout 4
with coefficients. Then the total energy error is sqrt(0.362 + (2*0
Rollout 14
]^2 (a/2 ) +3 sqrt(2 K a)(a/2 )2 +
Rollout 14
<= K/A.\n\nSimilarly, I1 <= sqrt( J1 ) * sqrt A.\n\nBut J1

DOWN_PROJ

Top 16 Positive Activations
Rollout 4
\pm 0.02 \mathrm{~V}$. What is the percentage error in the
Rollout 1
\frac{d}{\sqrt{2}} \).\n\n3. ** using Hilbert Space Properties**
Rollout 7
P = \frac{2N}{3} \) and \( R = \frac{N}{
Rollout 12
\sqrt{1 - \nu^2} \). Thus, we can express \( f(\nu
Rollout 1
\frac{d}{\sqrt{2}} \) can be non-empty due to the finite intersection
Rollout 12
\sqrt{1 - \nu^2} \). The derivative of \( K(k) \)
Rollout 1
\frac{d}{\sqrt{2}} \).\n - For any two distinct \( x,
Rollout 1
helpful mathematics assistant.\n<|im_start|>user\nLet $ \mathcal{H}$ be an infinite-dimensional
Rollout 7
\( R = \frac{N}{3} \), solving gives \( x = \frac{N
Rollout 12
) = \frac{1}{\nu} \). The integrand \( F(x, \nu
Rollout 12
1 - \nu^2 x^2)}} \). Let's apply the formula.\n\nFirst, compute
Rollout 12
) = \frac{1}{\nu} \), so \( b'(\nu) = -
Rollout 7
x = \frac{N}{12} \).\n - \( x \) must be integer
Rollout 12
\frac{1}{\nu^2} \).\n\nNow, evaluate \( F \) at \(
Rollout 13
{r}{n}\right)^{nt} \).\n\n**For annual compounding (n =
Rollout 1
answer\nGiven an infinite-dimensional Hilbert space \( \mathcal{H} \) and a set
Top 16 Negative Activations
Rollout 14
+ K/A.\n\nPlug A=2K/ sqrt(J):\n\nFirst term sqrt(J)* sqrt(A)=
Rollout 11
\boxed{(-1, 3)}.\n\n**Final Answer**\n\boxed{(-1,
Rollout 3
\boxed{169} m/s.\n\n**Final Answer**\nThe mean molecular speed of radon
Rollout 4
, 0.360583333... �� 0.00
Rollout 14
warz inequality with appropriately chosen vectors / weighet\n\n**Final Answer**\n\boxed{ \left(\int
Rollout 1
exists as described}}" or as per problem statement.\n\n**[Final Answer]**\n\boxed{\text{
Rollout 2
boxed answer is \boxed{109}\n\n**Final Answer**\n\boxed{109}\n\n
Rollout 4
.774 * 3.1623e-3 �� 8.7
Rollout 7
.\n\nTherefore, part(b) has no solutions.\n\n**Final Answer**\n\n(a) \boxed{12
Rollout 7
+5 >=65, that's original score >=60. Wait, but originally, the pass
Rollout 0
,3,5,7 are even?\n\nWait floor operations:\n\nFor prime 2: 20
Rollout 12
1.\n\nHence, the proof is complete.\n\n**Final Answer**\n\n\boxed{f(\nu)
Rollout 10
, the second player can win in these cases.\n\n**Final Answer**\n\boxed{There are infinitely many
Rollout 10
a way that leads to victory, those are called losing positions or P-positions in combinatorial game
Rollout 4
004)^2 )\n\nCalculating 2*0.004 = 0.0
Rollout 7
59.\n\nWait, but that suggests that perhaps the increase affects the promotion status? But the problem

Layer 13

GATE_PROJ

Top 16 Positive Activations
Rollout 12
[P. Turan]\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to prove that the
Rollout 1
mathcal{H}$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to show that there
Rollout 6
hash of α_n. Not sure.\n\nWait, think about P_angel: the original TM M is
Rollout 11
1/c = bc + ac + ab. So p can be written as (a + b + c
Rollout 14
dx \right) $$ \n<|im_start|>assistant\n<|im_start|>think\nOkay, so I have to prove this inequality
Rollout 11
) + (ab + bc + ca). So p = a + b + c + ab + bc
Rollout 9
Wait, hold on. \n\nWait, in the square AIME, let me confirm the positions. If
Rollout 6
$ . [/*]\n<|im_start|>assistant\n<|im_start|>think\nOkay, let me try to work through these
Rollout 11
ca) + 2. Wait, so then p = (1 + a)(1 + b)(
Rollout 11
.\n\n*David Altizio*\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I have this problem here where
Rollout 11
nπ/3.\n\nBut earlier, from the p condition, �� must satisfy cosθ=-2
Rollout 11
. That might be easier to work with. So p is equal to the sum of the variables plus the
Rollout 11
= 1/a. Then, it's a cyclic relationship. But let's test this.\n\nSuppose
Rollout 11
here's an important point: If we have complex t, then t +1/t is real if and
Rollout 5
/100$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. I need to
Rollout 11
/a is bc. Let me explore this. So p is a + b + c + bc + ac
Top 16 Negative Activations
Rollout 7
constraints missed. Alternatively, perhaps N=24?\n\nLet me check manually for N=24.\n\n
Rollout 3
69. Maybe 168 m/s.\n\nBut checking further on speed, considering the problem might
Rollout 0
, the formula is 2^{k -1).\n\nHence, for 20! with
Rollout 3
is approximately 168.5 m/s.\n\nIn absence of more specifications, but given standard calculations
Rollout 3
Alternatively, perhaps they have in mind 0°C.\n\nAlternatively, maybe the problem wants the root mean square
Rollout 3
decimal, 168.5 m/s.\n\nHowever, considering that we used approximate pi and calculator
Rollout 3
original exact formula gives 168.5.\n\nBut again, lacking temperature specification, this question is
Rollout 3
/s. Alternatively, maybe exact problem has specific answer.\n\nAlternatively, maybe the problem uses R=8.
Rollout 7
answer for part(b) is N=24?\n\nWait, wait. Earlier we saw that in equations
Rollout 3
. Alternatively, given that room temp is common too.\n\nBut since there's no specified temperature, usually,
Rollout 3
, which would be 161 m/s.\n\nAlternatively, in some contexts, STP is
Rollout 3
67.2 m/s. Another possible result.\n\nSo after spending an hour on this, the possible
Rollout 3
likely leading to approximately 169 m/s.\n\nBut wait, perhaps my calculator computation precision was a
Rollout 3
is taken as 300 K for approximation.\n\nWait, given that no temperature is given, perhaps
Rollout 3
8K) so ~169 m/s.\n\nAlternatively, if the user is from somewhere particular:
Rollout 4
C_eq is 1206 pF.\n\nSimilarly, the minimum C_eq: C1=

UP_PROJ

Top 16 Positive Activations
Rollout 2
1000. As997 terms fromk=3 tok=999.
Rollout 0
? No, that's not correct. Each must have all primes exponents from 20! For
Rollout 15
't bound by mod2. But when you track based on rules (lamp is ON tomorrow only if exactly
Rollout 2
09 and891 for each additional term fromk=3 onward.\n\nWait, but the real
Rollout 12
F(ν, ν) + [ boundary term from upper limit at ν + ε when changing ν ]
Rollout 2
mod1000), then each subsequent term fromk>=3 is also-1 mod10
Rollout 7
"Find all possible values of N", but with given constraints.\n\nBut suppose that in this problem, the
Rollout 11
system of equations. For example, ifq is something like s^2 - 2t or similar
Rollout 8
lengths of each block are variable (must be at least 1, obviously). But the key is that
Rollout 1
problem is over the Hilbert space, so thinking as in R^n, for which we know that even
Rollout 11
, though p and q are real. But given that coefficients s and t are sums like a +b
Rollout 0
-free? No, that's not correct. Each must have all primes exponents from 20!
Rollout 14
K=1, so the problem becomes: given that ��f(x)^2 dx =1 and
Rollout 1
only for the elements of S.\n\nTherefore, given that for all x ∈ S, ⟨x,
Rollout 1
, closed balls are weakly compact.\n\nBut given that spheres are closed and convex. Wait, the sphere
Rollout 1
can be formed that way. But the problem is given that such a set S exists, we need to
Top 16 Negative Activations
Rollout 1
vectors in $ \mathcal{H}$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I
Rollout 12
$ 0<\nu<1$ . [P. Turan]\n<|im_start|>assistant\n<|im_start|>
Rollout 12
is monotonically decreasing in the interval $ 0<\nu<1$ . [P
Rollout 7
participants is $N < 40$ . The passmark is fixed at $65
Rollout 1
helpful mathematics assistant.\n<|im_start|>user\nLet $ \mathcal{H}$ be an infinite-dimensional
Rollout 7
of the repeaters $56$ . \nHowever, due to an error in the wording of
Rollout 12
$ 0<\nu<1$ . [P. Turan]\n<|im_start|>assistant\n<|im_start|>think
Rollout 14
infty $$ Prove the following inequality. $$ \left(\int_0^\infty
Rollout 12
$ . [P. Turan]\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to prove
Rollout 2
$ is divided by $1000$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I
Rollout 11
+ b + c) + (1/a + 1/b + 1/c), and q as
Rollout 11
have absolute value $1$ . Define\n\begin{tabular}{c c c} $
Rollout 1
$ \mathcal{H}$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to show
Rollout 1
points (not necessarily countable) in $ \mathcal{H}$ such that the distance
Rollout 12
nu<1$ . [P. Turan]\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I
Rollout 6
that are of length $n$ . \n\n[list=1]\n [*] Given $k

DOWN_PROJ

Top 16 Positive Activations
Rollout 3
Let me make sure.\n\nAccording to Maxwell-Boltzmann distribution, the mean speed is the average of
Rollout 3
(average) speed derived from the Maxwell-Boltzmann distribution:\n\n\[\nv = \sqrt{\
Rollout 10
theory here.\n\nRecall by Legendre's three-square theorem, a number n is expressible as a
Rollout 3
8 K (unless stated otherwise), perhaps answer around \boxed{169} m/s. But
Rollout 14
HS=1 and RHS=4. Something's not adding up here. Wait, but the question states
Rollout 2
of-1. However,-1^{997}= -1. So product mod125
Rollout 11
2i sinθ:\n\n2i sinθ[ 2 cosθ - ( r3 +1
Rollout 11
1/3). But then complex cube roots complicates.\n\nMaybe look for substitution where terms repeat. For
Rollout 14
x)^2 dx = [1/(4 λ2 μ)] ��0^{1} (1
Rollout 12
ν2)u2), but that might complicate.\n\nAlternatively, perhaps a substitution that would turn into
Rollout 4
)/(a + b)^2 * Δa )2 + ( (a2)/(a + b)^
Rollout 14
(2 K a ) +a /2 )3 ≤8 K a2.\n\nLet’s denote that
Rollout 1
. But with S possibly uncountable, this isn't feasible.\n\nAlternatively, take a countable subset
Rollout 15
is even. Therefore, x_{t+1}[i] = 1 iff number of ON neighbors
Rollout 14
=1 and RHS=4. Something's not adding up here. Wait, but the question states to
Rollout 14
\n\nBut again, the integral of 1 over [0, ��) is infinite. So that
Top 16 Negative Activations
Rollout 14
a )(a/2 ) +3 (sqrt(2 K a ))(a2/4 )
Rollout 14
: K/A= K/(2K /sqrt( J )) )=sqrt( J )= /
Rollout 12
ν) / ( (1/ν) sqrt( (t2 - ν2)(1 - t
Rollout 14
} J^{1/4} + (sqrt( J ) )/2.\n\nBut we are to
Rollout 14
term K/A= K/(2K / sqrt( J} ))= sqrt( J)/2.\n\n
Rollout 1
two vectors would be sqrt( (d / sqrt(2))2 + (d / sqrt(2
Rollout 1
(x - y) ||·|| (sqrt(2}/d}{(x' - y}
Rollout 5
[150 ± 10*sqrt(725)] / 2 = 7
Rollout 5
2500)] / 2\n\nsqrt(72500) = sqrt(1
Rollout 1
distance each being d and radii d / sqrt(2}) would be non-empty? Not necessarily.
Rollout 14
cube we have:\n\nI^3 <= (sqrt( J) sqrt(A ) + K/A )^
Rollout 14
]^2 (a/2 ) +3 sqrt(2 K a)(a/2 )2 +
Rollout 1
sqrt(2))2 + (d / sqrt(2))2 ) = sqrt( d2/
Rollout 12
(nu) would be d/dnu K(sqrt(1 - nu2)). Let’s note that for
Rollout 14
= ��0^infty f/(sqrt(λ)) * sqrt(lambda ) dx ≤ [ �
Rollout 0
, a can vary from 1 up to sqrt(20!) if possible, but since 2

Layer 14

GATE_PROJ

Top 16 Positive Activations
Rollout 13
1936\n\nTherefore, A_annual = 6000 * 1.5
Rollout 7
50N/3 +59N/3=209N/3.\n\nSimilarly,
Rollout 13
:\n\nA_annual = 6000*(1 + 0.12)^4\n
Rollout 7
/3, so 150N/3 +59N/3=209
Rollout 13
0.12. So:\n\nA_annual = 6000*(1 + 0
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6\n\nSo 6000 * 0.57351936 =
Rollout 7
.\n\nNow, subtract 209N/3 from both sides:\n\n16x =71
Rollout 13
]\n\[\nA_{\text{annual}} = 6000 \times 1.
Rollout 7
N/12 +162N/12 =224N/12=
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interest is compounded annually.\n\nFor annual compounding:\nA_annual = 6000*(1
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= 16.\n\nSo A_quarterly = 6000*(1 + 0
Rollout 9
20y\n\nSo 20y = 100 => y = 5.\n\n
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10) / (2000 * (5000)) = (30,
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3 -14N=56N/3 -42N/3=14N
Rollout 7
=71N -209N/3= (213N -209
Rollout 7
9N/3=209N/3.\n\nSimilarly, the x terms:\n\n(75
Top 16 Negative Activations
Rollout 15
n=1=2^0. Let's think of our n= 2^k.\n\nFor
Rollout 0
/2 and N cannot come in pairs. So hence exponents for primes bigger than N/2 will
Rollout 2
which satisfies the first congruence.\n\nWait, hmm, that seems contradictory. Wait, no, according
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real. Not sure.\n\nEven this direction is not leading anywhere. Finally, think of the answer. Since
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as arrangements where there are two contiguous color blocks. Hence, such sequences are called "linear arrangements with two
Rollout 14
, perhaps using Hölder's for two functions. Think of the three-factor inequality.\n\nAlternatively if we consider
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*T2.\n\nIs there a relationship? Hmmm.\n\nMoreover, since abc =1, a *
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, Take q=3/2 and similarly. Hmm. Alternatively:\n\nAssume that a=2 and
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that in equations, but led to impossibility. Thus, perhaps the answer is different.\n\nAlternatively, maybe
Rollout 10
at n=25. However, let's think modulo4.\n\nSquares mod4 are 0
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as coprime factors, the way I'm thinking is that the assignment is of each prime as a
Rollout 10
would mean n=4 is winning.\n\nWait a moment. Let me write them down:\n\n0: losing
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must give it entirely to a or b, which leads to these coprime pairs. So seems that
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q, which are real.\n\nAnd the problem is from David Altizio, which suggests there is a clever
Rollout 8
I think in probability problems with colored cards, unless specified, we need to consider each card as distinguishable
Rollout 1
as the graph of a linear operator or something. Hmm.\n\nAnother approach. Assume H has an orthon

UP_PROJ

Top 16 Positive Activations
Rollout 1
problem is over the Hilbert space, so thinking as in R^n, for which we know that even
Rollout 11
bc+ ca. compute p2. Maybe that expands to something that can be combined with q, but
Rollout 1
equations intersect at a common y.\n\nBut the problem now is to show that the system of equations ||x
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from ν to1, yes.\n\nSo the problem is now f(ν) = ��_{ν
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have a < b or a > b, so then to get the (a,b) with a <
Rollout 9
into two parts of 5 units each. But also, since EM is vertical, G is somewhere to
Rollout 1
≠ j. But actually no, in the previous, if we take multiple points, we need to ensure
Rollout 9
bottom-right and M is top-right, so that would be the vertical side. Wait, no, in
Rollout 14
substitution or integration by parts.\n\nWait, another idea is to use the Cauchy-Schwarz inequality
Rollout 11
+ ca. compute p2. Maybe that expands to something that can be combined with q, but seems
Rollout 1
x'||2 for any x, x', so then from the above equation, Re⟨x - x
Rollout 9
/h)(x -10), which rearranges to y = (-5/h)(x -10
Rollout 1
x, x', so then from the above equation, Re⟨x - x', y⟩ =
Rollout 5
) = 2/3. \n\nSo setting up the equation: (b + 25)/(
Rollout 8
G2 would not. But in the problem's terms, Kathy is only concerned with the colors. That
Rollout 3
1 m/s. However, if the problem is from a US textbook, some might take STP as
Top 16 Negative Activations
Rollout 14
the maximum possible ratio is approximately 2.25/8≈0.28. Wait,
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k, then any initial state will reach zero after k steps.\n\nTherefore, if we can show that M
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mutual distances are d, and the spheres have radius d / sqrt(2). Then, 2r
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modeling as mod2 is invalid because lamps are not mod2. For example, lamps can be toggled
Rollout 6
are a helpful mathematics assistant.\n<|im_start|>user\nLet's say a language $L \subseteq \{
Rollout 1
d. For example, if two points centers are d apart, spheres with radii r1 and r
Rollout 2
2=997 terms, each contributing a -1.\n\nSo first compute 9*99
Rollout 12
\n\nBut t is expressed in terms of theta via t = sqrt( ν2 + (1 - ν
Rollout 8
.\n\nWait, k=2: two sequences per k, contributing 2400. Wait,
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system can have overlaps where states aren't bound by mod2. But when you track based on rules (
Rollout 14
2^{2/3}*3^{1/3}. Then 12^{2/3}=
Rollout 15
with M^k, all states would die after k steps. However, in the original problem's dynamics
Rollout 14
0 to1/μ: μ*(1/μ3)/3= 1/(3 μ2
Rollout 15
state based on the exact condition, which matches what modulo 2 would produce. Thus, dynamic is precisely
Rollout 1
centers equals d2, radii squares are each d2 / 2. So sum of radii
Rollout 1
' contains 0 and other points each at distance d from 0 and distance d apart.\n\nBut this

DOWN_PROJ

Top 16 Positive Activations
Rollout 3
square speed, which is related to temperature and molar mass. Wait, but the question specifically says "
Rollout 3
the temperature in Kelvin, and M is the molar mass. Then there's the average speed, which
Rollout 3
Kelvin,\n- \( M \) is the molar mass of radon in kg/mol.\n\n**Steps
Rollout 2
9 -1). Therefore, the product is ��_{k=1}^{999}
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/mol. Let's check once again.\n\nThe molar mass of radon. Let me confirm.\n\nRad
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1 atm. At STP, the molar volume is 22.4 liters, which
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positions can be infinitely constructed.\n\nTo make this argument precise is non-trivial. However, we can,
Rollout 3
since it's a monatomic gas, the molar mass would be approximately 222 g/mol
Rollout 3
, Radon is a noble gas. Its molar mass is, since its atomic number is 8
Rollout 6
if the oracle confirms its presence.\n\n**Answer:** \nThe sparse set \( S \) consists of all
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a weighted Hölder's inequality. \n\nLet me recall: in a weighted Hölder's inequality, for
Rollout 3
For example, for nitrogen, which has a molar mass of 28 g/mol, the average
Rollout 12
, k) = K(k), the complete elliptic integral of the first kind. Hence:\n\nTherefore,
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of power functions. But not sure. Let me recall the Beta function:\n\nB(p, q) =
Rollout 3
they just want the formula expressed in terms of molar mass and R? No, the question is to
Rollout 3
/s.\n\nAs an assistant, I need to perhaps suggest both possibilities, but likely assume STP. Alternatively
Top 16 Negative Activations
Rollout 3
the mean speed is called average speed and given by that formula with 8/π, while the rms
Rollout 1
d / sqrt(2) + d / sqrt(2) = d sqrt(2} ~1
Rollout 9
and GM.\n\nEquation for GE (from G(10 - h,5) to E(
Rollout 1
a sphere with center y and radius d / sqrt(2), then the necessary and sufficient condition is that
Rollout 5
oid into two smaller trapezoids? Each with height h/2, if the original height is
Rollout 1
a regular simplex in R^n with edge length sqrt(2), then the vectors from the centroid to the
Rollout 1
+ d / sqrt(2) = d sqrt(2} ~1.414 d >
Rollout 1
y } is orthogonal system with norm d / sqrt(2).\n\nFrom Rudin or another resource, there
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0? Wait, let's verify:\n\nFrom G(10 - h,5) to E(
Rollout 1
sqrt(2))2 + (d / sqrt(2))2 ) = sqrt( d2/
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equal to its right neighbor (since it has only that neighbor). Since the next state is on iff exactly
Rollout 1
centered at x ∈ S with radius d / sqrt(2)) is non-empty.\n\nBut how to check
Rollout 1
we need points at y's distance d / sqrt(2}, and vectors orthogonal.\n\nBut perhaps in this
Rollout 1
sum of radii l_s = d / sqrt(2) + d / sqrt(2) =
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to x and x' with distance d/sqrt(2), then the inner product is zero. So
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each lying on the sphere of radius d/sqrt(2) around y, and with pairwise inner product

Layer 15

GATE_PROJ

Top 16 Positive Activations
Rollout 1
are mutually orthogonal after scaling.\n\nWait, let me recall that if you have a regular simplex in R^n
Rollout 9
}\n\nWait, that seems correct.\n\nLet me check this.\n\nGiven that the area is 100
Rollout 5
the two bases. Wait, no, let me check. \n\nI remember that for areas divided by a
Rollout 15
mathematics assistant.\n<|im_start|>user\nLet $n\geq1$ be a positive integer.
Rollout 1
two distinct vectors are orthogonal.\n\nFirst, let me recall that an orthonormal system requires two things:
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{dx}{\sqrt{(x^2 - 1)(1 - \nu^2 x^
Rollout 14
with exponents that are conjugate. Let me recall Hölder's inequality: for $p$ and
Rollout 14
/(J K) becomes dimensionless? Let me check dimensions. Suppose f(x) is a function of
Rollout 1
bert space, let $ d>0$ , and suppose that $ S$ is a
Rollout 14
and 3/2.\n\nWait, let me think about the three-way Hölder: for three functions
Rollout 5
5). Is that right?\n\nWait, let me check the calculations again step by step. Probably, I
Rollout 14
Unless miscalculations.\n\nWait, let me check calculations again. \n\nFrom the beginning:\n\nGiven J
Rollout 15
1, there's only one lamp. Let me check.\n\nHmm, the problem says, "the two
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mod1000. But let me check this process again.\n\nFirst modulus8:\n\nProduct=(1
Rollout 14
a weighted Hölder's inequality. \n\nLet me recall: in a weighted Hölder's inequality, for
Rollout 4
0.88%. Wait, let me check the exact value.\n\nFirst, compute (ΔC
Top 16 Negative Activations
Rollout 2
;\n\nk=2,10^2 -1=99\n\nk=3, so1
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=3, which is 2^2 -1: no, since n=3 was bad.
Rollout 15
Because n=1 is 2^1 -1 =1, and then maybe 3,
Rollout 2
00\n\nSimilarly,10^k -1= -1 mod1000 fork>=
Rollout 15
the solutions. So, perhaps powers of two minus one? Because n=1 is 2^1
Rollout 15
, n=1 is 2^1 -1. Let's try n=3, which is
Rollout 2
k=3, so10^3 -1=999≡-1 mod10
Rollout 2
For k=1,10^1 -1=9;\n\nk=2,10^
Rollout 2
modulo 8 is \(2^k - 1\).\n - For \(k \geq
Rollout 2
9 is odd. 10^k-1 in general is written as (10-1
Rollout 2
8\), so \(2^k - 1 \equiv -1 \mod 8\).\n
Rollout 2
with k>=3 are10^k -1. For10^3=100
Rollout 2
00. Therefore,10^k -1= -1 mod1000 forall k
Rollout 0
. If the original rational number is in lowest terms, that means that a and b are copr
Rollout 2
-1). But (10^4 -1)=9999. 999
Rollout 11
then z + \overline{z} is real, and z \overline{z} is

UP_PROJ

Top 16 Positive Activations
Rollout 11
values of the ordered pair $(p,q)$ .\n\n*David Altizio*\n<|im_start|>assistant\n<|im_start|>think
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to $EM$ in $\triangle GEM$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I
Rollout 5
exceed $x^2/100$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's
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prime positive integers. Find $m + n$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's
Rollout 15
a helpful mathematics assistant.\n<|im_start|>user\nLet $n\geq1$ be a positive
Rollout 1
a helpful mathematics assistant.\n<|im_start|>user\nLet $ \mathcal{H}$ be an infinite
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helpful mathematics assistant.\n<|im_start|>user\nSuppose $a,\,b,$ and $c
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.\n<|im_start|>user\nLet's say a language $L \subseteq \{0,1\}
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$ 0<\nu<1$ . [P. Turan]\n<|im_start|>assistant\n<|im_start|>think
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$ is divided by $1000$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I
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helpful mathematics assistant.\n<|im_start|>user\nA function $f:[0,\infty)\to[0
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that are of length $n$ . \n\n[list=1]\n [*] Given $k
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product of different primes, and how that relates? For example, consider assigning the small primes to a
Rollout 1
\nLet $ \mathcal{H}$ be an infinite-dimensional Hilbert space, let
Rollout 7
of promoted becomes79. To achieve this, need their total score:\n\ntotal promoted score =79
Rollout 6
as required.\n\nObviously, in an exam setting, need to explain more.\n\nAnother angle: Let S_L
Top 16 Negative Activations
Rollout 11
1. Since 1 has modulus 1 and argument 0, so modulus r_a r_b r
Rollout 11
such that a = \overline{b}, c is real? But then c must not be real
Rollout 11
+1/r3.\n\nNote that if R = r3 +1/r3, then cosθ =
Rollout 11
consider all three variables with the same modulus, say r, but different arguments. Then since product is
Rollout 11
�, b = a (so same modulus and argument), c =1/a2. Then compute p
Rollout 11
for some integer k. Hmm, as modulus: r_a r_b r_c = 1.\n\nBut
Rollout 11
r_a e^{iθ_a}, b = r_b e^{iθ_b}, c = r
Rollout 11
1. Then, p = x + y + z + xy + yz + zx. And q
Rollout 14
=1?\n\nAlternatively, set J=1, K=1, so the problem becomes: given that
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+ bc + ca. Then P = S + T. )\n\nGiven that abc=1, Then inverse
Rollout 11
modulus 1 and argument 0, so modulus r_a r_b r_c = 1, and
Rollout 11
ab). Let me try that. Let's write c as 1/(ab). Then q becomes a
Rollout 11
, work with variables s = a + b + c, t = ab + bc + ca, which
Rollout 11
arguments sum to zero. Let’s pick a = r e^{iθ}, b = r e^{-
Rollout 11
But a and b have modulus r ≠1, c has modulus 1/(r2). Also,
Rollout 11
is 1. But a and b have modulus r ≠1, c has modulus 1/(r

DOWN_PROJ

Top 16 Positive Activations
Rollout 12
_a^b \frac{dt}{\sqrt{(t - a)(b - t)(t
Rollout 12
:\n\n∫_a^b \frac{dt}{\sqrt{(t - a)(b - t
Rollout 12
:\n\nIntegral from ν to 1 of dt / sqrt( (t2 - ν2)(1 -
Rollout 12
from ν to1. Suppose we set t = sqrt(ν2 + (1 - ν2)
Rollout 12
0 is ~ ��0 ... du / sqrt(u)\n\nWhich is convergent. Hence, the integral
Rollout 12
t=1 of (dt/ν) / sqrt( ( (t/ν)^2 -1
Rollout 12
�0^{pi/2} [1 / sqrt(1 - k2 sin2theta) ]
Rollout 4
696. The square root of that is sqrt(0.0000769
Rollout 11
. So for p to be real, p = conjugate(p). Similarly for q. So let's
Rollout 12
}^{π/2} [ dθ / sqrt( nu2 + (1 - nu2)
Rollout 14
\right) $$ \n<|im_start|>assistant\n<|im_start|>think\nOkay, so I have to prove this inequality:
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how his opponent plays.\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to prove that in this
Rollout 12
}^{pi/2} [ 1 / sqrt(1 - k2 cos2theta) ]
Rollout 14
2K / sqrt(J))^{1/2 }= sqrt(Jericho}( (2K))
Rollout 11
}) + (r e^{-iθ})(1/r2 ) + (1/r2)( r e
Rollout 1
cal{H}$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to show that there exists
Top 16 Negative Activations
Rollout 6
and computed deterministically, it's like a P machine.\n\nSo unless NP = P, which is unknown
Rollout 6
oly theorem. The connection between sparse oracles and P/poly is established that a language is in P
Rollout 6
in complexity theory that P/poly is exactly the class of languages decidable by a polynomial-time machine with
Rollout 6
A(n).\n\nSo P_bad-angel is the class of languages decidable by a poly-time machine with
Rollout 6
's theorem equivalence between \( \textbf{P}/\text{poly} \) and sparse or
Rollout 13
1.25 = 1.5625. Then, 1.254
Rollout 3
, standard temperature for gas calculations is often 0°C, which is 273.15
Rollout 6
machine with a sparse oracle. (This is the Meyer, I think, theorem.) So perhaps this is
Rollout 14
exponents 2 and 2 gives Cauchy-Schwarz, which we can relate J and
Rollout 12
/2, k) = K(k), the complete elliptic integral of the first kind. Hence:\n\n
Rollout 12
not sure. Let me recall the Beta function:\n\nB(p, q) = ��_0^
Rollout 13
25)^2 = 1.5625\n\n2*1.25*0
Rollout 6
, recall that P_angel is equivalent to P/poly, but with the advice strings being generated non
Rollout 14
itty with K.\n\nAlternatively, start with Cauchy-Schwarz:\n\nI = ��0^
Rollout 6
^*$ is in $\textbf{P}_{angel}$ if there exists a polynomial
Rollout 6
oracle S_L. It is a standard result in complexity theory that P/poly is exactly the class of

Layer 16

GATE_PROJ

Top 16 Positive Activations
Rollout 11
the same modulus.\n\nAlternatively, two variables with reciprocal modulus and one with modulus not 1.\n\nBut overall
Rollout 11
= \overline{1/b}, leading to modulus |a| =1/|b | ;
Rollout 11
real. But since a is neither real nor of modulus 1, this is not the case.\n\nWait
Rollout 11
|=1/k2 <1. So none have modulus 1, which is okay. Then, set
Rollout 11
= ��, b = a (so same modulus and argument), c =1/a2. Then
Rollout 11
1. Let’s consider them with moduli reciprocal-related. Like |a|=k>1,
Rollout 11
?\n\nAlternatively, recall the problem states that none have modulus 1. If we let’s take a,
Rollout 11
. Hence, all three variables cannot have the same modulus.\n\nAlternatively, two variables with reciprocal modulus and one
Rollout 11
1. But problem states that none of them have modulus 1. Hence, |a| and |
Rollout 11
modulus 1. Let’s consider them with moduli reciprocal-related. Like |a|=k>1
Rollout 11
product is 1. But a and b have modulus r ≠1, c has modulus 1/(
Rollout 11
}, c= e^{-i2θ}. But modulus of a and b is 1, which is
Rollout 11
none of a, b, c are real or modulus 1. Let’s consider them with moduli
Rollout 11
their product is 1, so if two have modulus greater than 1 and one less, but their
Rollout 11
�, cube roots of unity. But they have modulus 1. Not allowed. Problem explicitly says modulus
Rollout 11
, setting variables in terms of exponentials. Since modulus isn't 1, maybe log variables?\n\nBut
Top 16 Negative Activations
Rollout 1
2 - 3d2/4 - 3d2/4 + d2 / 2
Rollout 11
2i sin2θ = 2i*2 sinθ cosθ =4i sinθ cos
Rollout 1
x'⟩ = 2d2 - 2⟨x, x'⟩.\n\nTherefore:\n\n⟨
Rollout 1
2 = 2||u||2 + 2||v||2.\n\nGiven that S is a
Rollout 1
0 ⇒ (2c - d2 - 2c + d2)/2 = 0 ⇒
Rollout 14
^{1/3}= (2^2 *3 )^{1/3}=2^{2/
Rollout 14
.\n\nNote that 12=22 *3, so 12^{1/3}=
Rollout 11
2*(2 e^{iθ}) + (4 e^{i2θ}) + (1/
Rollout 1
equations,\n\n2⟨x, y⟩ - 2⟨x', y⟩ = 2⟨x
Rollout 11
= 4 e^{-iθ} + 4 e^{-i2θ} + (1/
Rollout 4
4)^2 )\n\nCalculating 2*0.004 = 0.008
Rollout 11
= r^4 +2r2 + 2/r2 +1/r^4\n\nBut we
Rollout 11
= 4 e^{-iθ} + 4 e^{-i2θ} + (1/
Rollout 1
(2c - d2)/2 - (2α - ||y||2) = 0
Rollout 2
3, and 10 is 2*5, so 10 and 125
Rollout 13
}{4}\right)^{4 \cdot 4} = 6000 \left(

UP_PROJ

Top 16 Positive Activations
Rollout 3
that standard problems involving heavy gases like radon typically compute the speed under standard conditions, either 0°C
Rollout 15
1} = M x_t, where M is such that for each lamp, it's the sum of
Rollout 3
gas experiments, diffusion rates, Graham's law, typically at STP. But the molar mass is
Rollout 10
each move is subtract a square, so perhaps using octal games concept or mathematical induction.\n\nAlternatively, is
Rollout 3
170? Due to perhaps answer key conventions. For purposes of being precise. If allowed as
Rollout 0
a subset S of the prime factors of N, hence product(a)= product of primes in S (with
Rollout 11
. But since r is a positive real number ( modulus ), r ≠1. Also, modulus is related
Rollout 3
whether to use degrees Celsius or Kelvin. Wait, no, temperature must be in Kelvin in the formulae
Rollout 6
can have each block indexed by n and position, hence for each n, the number of strings is O
Rollout 14
infty x f(x) dx$ becomes a first moment. Maybe use scaling. Let’s set $
Rollout 3
15 K. Let me check.\n\nWait, possibly, the problem expects STP conditions (Standard Temperature
Rollout 4
and find the difference. But given that the problem provides the uncertainties as ± values, I think the standard
Rollout 8
Wait, wait, no. Wait, the problem says "shuffles the 10 cards and lays
Rollout 8
) but considering order.\n\nWait, but when they compute the probability, is the sample space the number of
Rollout 15
no neighbors. Wait, hold on, the problem says "the two lamps at the ends have one neighbor
Rollout 11
Let’s pick specific values. Wait, the problem says that a, b, c can't be on
Top 16 Negative Activations
Rollout 0
are related to the divisors of 20! where each pair consists of two coprime numbers
Rollout 5
25, greatest integer not exceeding 181.25 is 181. Therefore
Rollout 11
2 e^{i2θ}). Product a*b*c = r2 e^{i2θ} *
Rollout 0
prime powers in the factorization of 20! must go entirely to one of the numbers \(a
Rollout 12
) ) / t ] dθ\n\nTherefore, dt / [ sqrt( (t2 - ν2
Rollout 11
^{-i2θ}) + (e^{-iθ} -4 e^{-iθ}) =
Rollout 0
of ways to write 20! as a product of two coprime numbers, \(a\)
Rollout 11
i2θ} - 3 e^{-iθ} = 0\n\nWhich can be expressed in
Rollout 12
t\n\nSo, rearranged, the integrand dt / sqrt(...) becomes dθ / t\n\nBut
Rollout 0
!. So they form a coprime pair whose product is 20!. Such pairs are related to
Rollout 5
\]\nThe greatest integer not exceeding 181.25 is 181.\n\n\
Rollout 0
the others are even. Thus, 20! is not a square because it has primes raised to
Rollout 0
rime, each prime factor in 20! must go entirely to \(a\) or entirely to
Rollout 0
:\n - The prime factors of 20! are 2, 3, 5,
Rollout 0
are even and odd.\n\nTherefore, 20! is not a perfect square because, for example,
Rollout 12
1/2}. Hence, near t=ν, F(t, ν) is of order (t

DOWN_PROJ

Top 16 Positive Activations
Rollout 12
int_1^{\frac{1}{\nu}} \frac{dx}{\sqrt{(x
Rollout 12
we had x from 1 to 1/ν, substitution t=νx gives t from ν
Rollout 12
1/\nu \):\n\n\( F(1/\nu, \nu) = \frac{1}{
Rollout 12
x = 1 to x = 1/ν of dx over sqrt( (x2 -1
Rollout 10
't work.\n\nPossible idea: The sequence of Grundy numbers must eventually be periodic due to the finite number
Rollout 10
numbers or Nimbers. Each position has a Grundy number, which is the mex (minimum exclud
Rollout 12
νx. Then when x = 1/ν, t = 1, and when x =
Rollout 14
,\infty)$ is integrable and $$ \int_0^\infty f(x)^
Rollout 1
the equations are consistent.\n\nHence, the problem reduces to finding a t in H such that for all
Rollout 2
? 10 and 125 share factors: 5. Hmm. 125
Rollout 12
at the upper limit \( x = 1/\nu \), multiplied by the derivative of the upper limit
Rollout 10
mex (minimum excludant) of the Grundy numbers of positions reachable from it. A position is
Rollout 10
at n=25.\n - The Grundy numbers (mex values) were computed for several
Rollout 12
F \) at \( x = 1/\nu \):\n\n\( F(1/\nu, \
Rollout 1
for all x ∈ S.\n\nThus, the problem reduces to finding y such that the function f(x)
Rollout 10
<=n}.\n\nSo let's compute the Grundy numbers:\n\nInitialize Grundy[0]=0.\n\n
Top 16 Negative Activations
Rollout 14
c - dx)^2= c^2 -2c d x + d^2 x^2
Rollout 7
) + (N/12)(A -79)=0\n\nSimplify:\n\n-2N
Rollout 3
or 298 K? Wait, standard temperature for gas calculations is often 0°C, which
Rollout 7
.\n\nTherefore:\n\n(N/12)(A -54) = 2N/3\n\nDiv
Rollout 3
6.512*(300 -7) = 66.512*
Rollout 2
25 mod8: 125 /8= 15*8=120
Rollout 12
K(k) ] / (k(1 -k2))\n\nTherefore, given f(nu) =
Rollout 12
<0, since the denominator is 1 -nu2>0 in the interval (0,1
Rollout 12
nu2, so that denominator k(1 -k2) becomes k nu2.\n\nThus,\n\nd
Rollout 12
- nu2), so k2 =1 -nu2\n\nThus:\n\nf’(nu) =
Rollout 13
+ b)^2 = a^2 + 2ab + b^2.\n\nPerhaps set a =
Rollout 3
possibly, the problem expects STP conditions (Standard Temperature and Pressure), which is 0°C and
Rollout 7
N/3\n\nFactor x:\n\nx(A -54) +18N =56N
Rollout 8
40=744/3024=744÷24=31
Rollout 12
) K(k)) / (k(1 -k2))\n\nWait let me verify:\n\nK(k)
Rollout 3
Thus, gas at STP. Then STP temperature is 273 K. So maybe they

Layer 17

GATE_PROJ

Top 16 Positive Activations
Rollout 3
(3RT/M), where R is the gas constant, T is the temperature in Kelvin, and M
Rollout 3
where:\n- \( R \) is the gas constant (\(8.314 \, \
Rollout 3
222 kg/mol.\n\nR is the gas constant, which is 8.314 J
Rollout 3
, maybe they expect us to recall the molar mass of radon and the temperature at STP.\n\n
Rollout 3
speed formula is sqrt(3RT/M), where R is the gas constant, T is the temperature in
Rollout 4
mathrm{~V}$. What is the percentage error in the calculation of the energy stored in this combination
Rollout 3
, textbook's choice. Given a chemistry book, Oxtoby, example problem: At 25
Rollout 3
if for some reason someone mixed them up.\n\nRMS speed for Rn at 298 K
Rollout 3
It's tricky.\n\nWait, given that molar mass is 222 (3 sig figs
Rollout 3
,\n- \( M \) is the molar mass of radon in kg/mol.\n\n**Steps:
Rollout 4
2 *10, etc. Let's recalculate the derivatives values more precisely:\n\nStarting with ΔC
Rollout 3
temperature in Kelvin, and M is the molar mass. Then there's the average speed, which is
Rollout 3
1 atm. At STP, the molar volume is 22.4 liters, which is
Rollout 3
, typically at STP. But the molar mass is key. So radon is often used in
Rollout 3
. Let me see. Let's do a quick estimation. Let's take 298 K first
Rollout 3
precise calculation using more decimal places.\n\nLet me recalculate the exact value.\n\n8 * 8.3
Top 16 Negative Activations
Rollout 10
n ��0 mod4: subtract0 or1.\n\n subtract s≡0 or1 mod4,
Rollout 10
subtract0 or1.\n\n subtract s≡0 or1 mod4, n -s≡0 -0
Rollout 10
)\n\nBut since squares modulo4 are 0 or1. Therefore,n being0,1,2,
Rollout 0
of 20! must go entirely to one of the numbers \(a\) or \(b\).
Rollout 15
i, next state is (left neighbor state) XOR (right neighbor state). If lamp i is in
Rollout 15
is both ends, so does it have zero neighbors or one? Wait, the problem statement says "the
Rollout 10
that all n > some number are either in S or can reach S by subtracting a square. Which
Rollout 9
: y=(-1/5)(10)+2= -2 +2=0, GM:
Rollout 10
4.\n\nSquares mod4 are 0 or1. Hence subtracting square can make the residues:\n\n
Rollout 10
If n ��0 mod4: subtract0 or1.\n\n subtract s≡0 or1 mod4
Rollout 10
n ��1 mod4: subtract1 or0.\n\nn -s≡0 or1.\n\nSimilarly
Rollout 10
1 or0.\n\nn -s≡0 or1.\n\nSimilarly, maybe analysis similar to NIM in
Rollout 10
: subtract0 or1.\n\n subtract s≡0 or1 mod4, n -s≡0 -
Rollout 10
my initial conclusion that losing positions are n≡0 or2 mod5 was correct up to certain n but
Rollout 0
prime factor must go either to a or to b, since a and b are coprime. Therefore
Rollout 0
a,b) with a*b = N is equivalent to choosing a subset S of the prime factors of N

UP_PROJ

Top 16 Positive Activations
Rollout 14
.31 which cubes to ~2.25, which is way small than8, indicating that
Rollout 14
4. Then divide by 4≈1.31, as above. So according to this,
Rollout 1
sqrt(2) d �� 1.414d. The mutual distance between centers is
Rollout 3
\n\nWait no, perhaps with precise calculation using more decimal places.\n\nLet me recalculate the exact value.\n\n
Rollout 14
and K=1 is around 2.25, which is less than 8. Therefore,
Rollout 7
might have been misconsidered.\n\nWait, pass mark is fixed at65. Originally, promoted are
Rollout 14
(9/4), which is approx1.31. But then the inequality requires I^3
Rollout 14
4)^{1/3}≈1.31... Wait, but that contradicts our previous
Rollout 14
^{2/3}/4 ��1.31 which cubes to ~2.25,
Rollout 3
/s}\n\]\n\n4. **Rounding to Significant Figures:**\n Given the inputs (molar
Rollout 3
For purposes of being precise. If allowed as per rounding rules, 168.5 rounds to
Rollout 14
4= 9/4=2.25.\n\nThen 8 J K=8*1
Rollout 14
the maximum possible ratio is approximately 2.25/8≈0.28. Wait,
Rollout 14
1.31)^3≈2.25. But 8 J K=8*1
Rollout 4
) * Δb )^2 ]\n\nPlugging numbers:\n\na =2000, b=
Rollout 3
Wait no, perhaps with precise calculation using more decimal places.\n\nLet me recalculate the exact value.\n\n8
Top 16 Negative Activations
Rollout 14
2 ) +3 (sqrt(2 K a ))(a2/4 ) +a3/
Rollout 11
2θ}) + (e^{-iθ} -4 e^{-iθ}) = 0\n\n
Rollout 11
})\n- ab = (r e^{iθ})2 = r2 e^{i2θ}\n
Rollout 11
2θ}) + (r e^{iθ})2 + (r e^{iθ})(1
Rollout 11
:\n\na/b = (r e^{iθ})/( r e^{-iθ} )) = e
Rollout 11
i2θ}))*(r e^{iθ}) = (1/r) e^{-iθ}\n\n
Rollout 11
(1/r2)( r e^{iθ}) ) = r2 + (1/r) e
Rollout 11
a^3 is (2 e^{iθ})^3 =8 e^{i3θ},
Rollout 14
(√(2 K J^{1/2 })) + (�� J)/ 2.\n\nLet me
Rollout 11
}\n\nb/c = (r e^{-iθ})/(1/r2 ) = r3 e^{-
Rollout 7
Similarly, the x terms:\n\n(75 -59)x=16x.\n\nTherefore, the
Rollout 11
p = 2*(2 e^{iθ}) + (4 e^{i2θ}) +
Rollout 14
[∫0^1 (1 -2 t + t2 )dt ] = [1/(4
Rollout 14
A^∞ (f(x) \sqrt{x}) \cdot (1/\sqrt{x}) dx.\n\n
Rollout 14
) +3 (sqrt(2 K a ))(a2/4 ) +a3/8
Rollout 10
uent to 0 or 2 mod 5? Wait: 0 mod5=0,

DOWN_PROJ

Top 16 Positive Activations
Rollout 3
(3RT/M), where R is the gas constant, T is the temperature in Kelvin, and M
Rollout 0
in N's prime factorization), which requires multiplicativity. For each prime p | N, assign all
Rollout 3
222 kg/mol.\n\nR is the gas constant, which is 8.314 J
Rollout 3
where:\n- \( R \) is the gas constant (\(8.314 \, \
Rollout 10
theory here.\n\nRecall by Legendre's three-square theorem, a number n is expressible as a
Rollout 12
of the complete elliptic integrals are:\n\ndK/dk = [ E(k) - (1
Rollout 1
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet $ \
Rollout 2
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nFind the remainder when
Rollout 6
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet's say a
Rollout 3
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nCompute the mean molecular
Rollout 10
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nConsider the following two
Rollout 7
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nIn a mathematics test
Rollout 4
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nTwo capacitors with
Rollout 8
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nKathy has $
Rollout 11
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nSuppose $
Rollout 9
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nSquare $AIME
Top 16 Negative Activations
Rollout 10
victory, those are called losing positions or P-positions in combinatorial game theory. The problem states
Rollout 10
10, the losing positions (P-positions) are 0, 2, 5
Rollout 10
next player can force a win) or P-positions (previous player can force a win). The key
Rollout 14
a -2)}.\n\nTake reciprocals on both sides:\n\nJ^{1/(2a -1
Rollout 10
otic density zero, then the set of P-positions can still be infinite. The intuition being that when
Rollout 10
), positions can often be classified as either N-positions (next player can force a win) or P
Rollout 10
n=2 is a losing position, so P-position.\n\nWait, contradicts my previous thought. Wait
Rollout 10
we need to find an infinite family of P-positions (Grundy number0). Since, even though
Rollout 10
The key is to identify losing positions (P-positions) where every move leads to a winning position (
Rollout 10
moves available), positions can often be classified as either N-positions (next player can force a win)
Rollout 7
5P = 10R\n\nDivide both sides by 5:\n\nP = 2R
Rollout 5
)] = 125H\n\nDivide both sides by H (assuming H ≠ 0):\n\n
Rollout 5
150; subtract 2b from both sides: b + 75 = 1
Rollout 8
.\n\nSimplify this fraction.\n\nFirst, divide numerator and denominator by 240: numerator becomes
Rollout 10
way that leads to victory, those are called losing positions or P-positions in combinatorial game theory
Rollout 2
1)=109,000 -109=108,891

Layer 18

GATE_PROJ

Top 16 Positive Activations
Rollout 5
be b, then the longer base is b + 100. The midline length is (
Rollout 12
{dx}{\sqrt{(x^2 - 1)(1 - \nu^2 x^
Rollout 3
24.942*298 ��24.942*3
Rollout 5
b + 50) + (b + 100))/2 * (h/2
Rollout 12
{1}{\sqrt{(x^2 - 1)(1 - \nu^2 x^
Rollout 3
66.512 * 298:\n\nLet me compute this multiplication.\n\n66.
Rollout 11
1/c = ab. So 1/a + 1/b + 1/c = bc + ac
Rollout 5
is (b + 25)/(b + 75) = 2/3. \n\n
Rollout 3
8*8.314*298= about 19,820 (
Rollout 3
66.512 * 298 = Let me compute 66.51
Rollout 2
since 1000 is 8 * 125 and 8 and 12
Rollout 3
8.314 * 298\n\nFirst, compute 8*8.3
Rollout 3
66.512*298 �� 19,820.
Rollout 4
C1 / C1 = 10 / 2000 = 0.00
Rollout 3
66.512 * 298 = let's compute 66.51
Rollout 3
14 (4 sf) * 298 (3 sf) / 0.22
Top 16 Negative Activations
Rollout 1
$ is a set of points (not necessarily countable) in $ \mathcal{H
Rollout 1
is a set of points (not necessarily countable) in $ \mathcal{H}$
Rollout 10
of numbers which cannot be represented as sum of an even number of squares. Wait, sum of squares theorem
Rollout 9
0 (base EM) and two lines intercepting x=0. Probably the overlap inside is the tr
Rollout 6
x of length n, needs to get the bits of α_n, which is p(n) long,
Rollout 15
system, the problem reduces to determining for which n the associated matrix is nilpotent. If M^
Rollout 6
length of each string is O(log n + log i). Since p(n) is polynomial in n,
Rollout 15
and only if the number of lamps \( n \) is a power of 2. This result is
Rollout 15
k} for integers \( k \geq 0 \).\n\nBut wait, earlier examples suggest otherwise.\n\n
Rollout 15
system evolves to zero in GF(2) after k steps. Since dynamically, over the lamp problem,
Rollout 11
b |^2 real, i.e., real positive. However c=1/(|b |2
Rollout 15
have one neighbor). The problem asks for which n the cellular automaton is nilpotent. It's
Rollout 9
, GM intersects the left edge at (0,10 -50/h). Similarly, these are
Rollout 9
from (0,0) to (0,10)). Since G is to the left of the
Rollout 2
the number of such terms, if it's odd number, it's109, else89
Rollout 15
is nilpotent if and only if n is a power of two.\n\nHowever, in our case n

UP_PROJ

Top 16 Positive Activations
Rollout 0
2.\n\nWait, 7^2 divides into 20! but 7^3 does
Rollout 8
4=31; 3024÷24=126. So yes,
Rollout 8
3, etc. 126 is 2* 63=2*7*
Rollout 8
=31, 30240÷240=126. As above
Rollout 11
, if p = s + t is real, so the polynomial x^3 - s x^2
Rollout 2
125. 1000 divided by125 is8, so 10
Rollout 2
5. Hmm. 125 is 5^3, and 10 is
Rollout 3
square root of 28,408.12:\n\nsqrt(28,40
Rollout 4
8%. Since 0.877 is 0.88 when rounded to two decimal places
Rollout 8
1; 3024÷24=126. So yes, 31
Rollout 0
of the total pairs, so 256/2 =128. Hence 12
Rollout 8
240: 7440 divided by 240. Let's see:
Rollout 8
0240. 7440/30240= Let's check by
Rollout 4
compute that more accurately.\n\n4.327 / 1200. Let me calculate
Rollout 15
1]); since 1 XOR 1 =0, 0 XOR1=1, etc. However
Rollout 7
294. Which is 294/3=98 per student, but it's
Top 16 Negative Activations
Rollout 4
are independent and random, and combines them in quadrature, while the max-min approach assumes that all errors
Rollout 11
r3 +1/r3.\n\nNote that if R = r3 +1/r3, then cos
Rollout 4
is about ±1.3%, whereas the quadrature gives ~0.88%, the answer would
Rollout 5
solve this problem about a trapezoid with bases differing by 100 units. The segment
Rollout 11
of variables and their pairwise products. q is the cyclic sum of variables divided by each other. Hmm,
Rollout 5
height k creates a smaller trapezoid with bases a and x and height k. The ratio of
Rollout 9
. But that seems like trapezoid with bases at x=0 (the two points: (
Rollout 4
then compute the maximum and minimum possible energy considering the uncertainties and find the percentage difference. But error propagation using
Rollout 11
. Then, since abc = 1, the constant term would be -1 (for cubic x^
Rollout 9
lines is the area of trapezoid with bases at x=0 (length 8-2
Rollout 4
errors are independent and random, and combines them in quadrature, while the max-min approach assumes that all
Rollout 11
real coefficients? Because then their imaginary parts come in conjugate pairs. But the problem says that none are
Rollout 3
average) speed derived from the Maxwell-Boltzmann distribution:\n\n\[\nv = \sqrt{\frac
Rollout 14
I make a substitution x= cy where c is a constant. Consider f(x)=d g(y).
Rollout 11
, the coefficient of x is t, and the constant term is -1.\n\nBut given that a,
Rollout 3
. Let me make sure.\n\nAccording to Maxwell-Boltzmann distribution, the mean speed is the average

DOWN_PROJ

Top 16 Positive Activations
Rollout 2
)...(10^{999} -1). We need to compute P mod 10
Rollout 0
a = product_{p in S} p^{e_p}, and b = 20! /
Rollout 2
891*(1000 -1)=891,000 -8
Rollout 4
C1 / C1 = 10 / 2000 = 0.00
Rollout 2
is k=1 (10^1 -1), the second is k=2 (10
Rollout 2
=891*(1000 -1)=891*1000 -
Rollout 0
, with three primes, 2^(3 -1)=4, which matches what we saw for
Rollout 11
not sure. Alternatively, p is (a + 1/a) + (b + 1/b
Rollout 1
cal{H}$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to show that there exists
Rollout 5
be b, then the longer base is b + 100. The midline length is (
Rollout 0
19). Hence, 2^{8 -1}= 128. Hence, result.
Rollout 13
--------\n224 (1.12 * 2)\n1120 (1.1
Rollout 8
=1: [5! / (5 -1)! ] = 5! / 4!
Rollout 2
99 (10^999 -1). Therefore, the product is ��_{k
Rollout 14
to show that $I^3 \leq 8 J K$.\n\nHmm. So, the
Rollout 8
the number of possible sequences is 10 * 9 * 8 * 7 * 6
Top 16 Negative Activations
Rollout 11
equations. But without real roots. But from Descarte’s signs, if the polynomial with real coefficients has
Rollout 12
Wait, let's recall that when we use differentiation under the integral sign, as long as the integral conver
Rollout 11
terms of exponentials. Since modulus isn't 1, maybe log variables?\n\nBut let's assume for
Rollout 8
7440/30240= Let's check by dividing numerator and denominator by
Rollout 8
7440/30240=744/3024=
Rollout 12
the integrand is sufficiently smooth, we can differentiate under the integral sign even with variable limits. However,
Rollout 12
So the derivative f’(ν):\n\nUsing differentiation under the integral sign when limits also vary, with the
Rollout 12
apply Leibniz?\n\nHmm. Maybe differentiation under the integral sign is still possible here. Let me
Rollout 5
.5H.\n\nSo we have two equations:\n\n1. (75 + x)h =
Rollout 8
the number of ordered arrangements. For example, for R^r G^{5−r}: we need
Rollout 12
f’(ν) would include two contributions:\n\n1. The lower limit ν, leading to -F
Rollout 9
/h <5, so (0,50/h) is below (0,5).\n\n10
Rollout 11
with a specific example where both p and q are real. Let's see if such examples exist.\n\nExample
Rollout 8
6.\n\nTherefore, 31/126. Then check if they can be simplified. Since
Rollout 8
As above. Therefore 31/126.\n\nBut I wonder, was that correct? Let
Rollout 14
2)\n\nNow, we have two equations:\n\n(1) λ2 μ=1/12\n\n

Layer 19

GATE_PROJ

Top 16 Positive Activations
Rollout 11
,b,$ and $c$ are three complex numbers with product $1$ .
Rollout 14
2 λ=1/12→λ3=1/12→λ=(1/1
Rollout 1
a helpful mathematics assistant.\n<|im_start|>user\nLet $ \mathcal{H}$ be an infinite
Rollout 15
a helpful mathematics assistant.\n<|im_start|>user\nLet $n\geq1$ be a positive
Rollout 14
helpful mathematics assistant.\n<|im_start|>user\nA function $f:[0,\infty)\to[0
Rollout 14
_0^\infty f(x)^2 dx<\infty,\quad \int_0
Rollout 11
a,\,b,$ and $c$ are three complex numbers with product $1
Rollout 13
on the loan is $12\%$. She makes no payments for 4 years, but
Rollout 14
(\int_0^\infty f(x) dx \right)^3 \leq 8\
Rollout 9
helpful mathematics assistant.\n<|im_start|>user\nSquare $AIME$ has sides of length $10$ units
Rollout 14
\int_0^\infty xf(x) dx <\infty $$ Prove the following
Rollout 9
mathematics assistant.\n<|im_start|>user\nSquare $AIME$ has sides of length $10$ units.
Rollout 12
1^{\frac{1}{\nu}} \frac{dx}{\sqrt{(x^2
Rollout 2
times 999 \times \cdots \times \underbrace{99\cdots
Rollout 11
prohibited if it's a root of x^3 -1. Since |ω|=1. Hence
Rollout 14
1} ��0^∞ f(y) dy = λ^{a -1} I.\n\nJ
Top 16 Negative Activations
Rollout 11
: a/b is the same; b divided by 1/(ab) is b * ab = a
Rollout 4
�0.88%, which is half of 1.76% (max-min). But
Rollout 11
not sure. Alternatively, p is (a + 1/a) + (b + 1/b
Rollout 15
. For two neighbors, sum is 0 mod2 iff sum is even. Therefore, x_{t
Rollout 1
sphere has radius d / sqrt(2} approx 0.707d. So sum of
Rollout 15
, sum is 2 ��0 mod 2, so the lamp is off. If exactly one
Rollout 1
that for any x, ||v_x|| = 1, and for any x ≠ x', ⟨
Rollout 1
⟨v_x, v_{x'}⟩ = 0.\n\nCalculating the norm of v_x:
Rollout 12
, the integral itself is convergent at x = 1/ν, so the derivative might still be
Rollout 8
the number of possible sequences is 10 * 9 * 8 * 7 * 6
Rollout 14
= e^{-k x}$, with k > 0. Then compute the integrals:\n\n$I =
Rollout 7
is 75. Wait, that’s not 76. So even though original promoted had their
Rollout 12
{1}{\sqrt{(x^2 - 1)(1 - \nu^2 x^
Rollout 2
k -1 is odd. Because even minus1 is odd. Therefore, every term in the product is
Rollout 4
606% as relative error.\n\nWait, 0.36058333
Rollout 8
each green card identical? Because if that's the case, then a sequence is determined by the colors.

UP_PROJ

Top 16 Positive Activations
Rollout 12
t2) ) ~ 1 / sqrt(2ν(t - ν)(1 - ν2)
Rollout 12
integrand becomes ~ 1 / sqrt( 2ν u (1 - ν2) ), so
Rollout 12
the lower limit is ν, upper limit is1, and integrand depends on ν.\n\nSo by Le
Rollout 9
, y). But the triangle being isosceles with base EM. Since triangle GEM is is
Rollout 5
(18125) = sqrt(25 * 725) = 5
Rollout 14
itstep by step:\n\nsqrt(A)=sqrt(2K/sqrt( J ))}= (2K
Rollout 9
5)\) because the triangle is isosceles with \(GE = GM\).\n\n4. **
Rollout 9
y. But as the triangle is isosceles with GE = GM.\n\nCalculating GE distance:\n\n
Rollout 9
EM. Since triangle GEM is isosceles, meaning legs GE and GM are equal. Wait
Rollout 14
a/2.\n\nSo I <= sqrt( 2 K a ) + a/2.\n\nNeed to
Rollout 12
ivative term would be zero), lower limit isν, so its contribution is -F(ν, ν
Rollout 15
equivalent to the sum mod2 but in GF(1), but actually, the rule is exactly the sum
Rollout 12
is small, i.e., near nu ��1.\n\nOn the other hand, at k=1
Rollout 9
equal. Wait, if it's isosceles with base EM, then sides GE and GM must
Rollout 14
sqrt(A ) + K/A.\n\nPlug A=2K/ sqrt(J):\n\nFirst term sqrt(J)*
Rollout 14
a/2.\n\nNeed to show ( sqrt(2 K a ) +a /2 )3 ≤
Top 16 Negative Activations
Rollout 2
remainder when the product of numbers like 9, 99, 999, and so
Rollout 10
\n\nWait, okay, so from 0 to 10, the losing positions (P-positions
Rollout 5
segment connecting the midpoints of the legs divides the trapezoid into two regions with areas in the
Rollout 0
is some fraction as part of between 0 and 1. When written in lowest terms, numerator a
Rollout 5
which is the length of the segment that divides the trapezoid into two equal areas, and then
Rollout 8
5, those could be anywhere from 0 to 5 reds. Wait, but in the problem
Rollout 8
GG, etc. But something like RRRGR wouldn't work because the reds are not all together
Rollout 0
Let's ask for rationals between 0 and 1 with product a*b=720,
Rollout 11
+ b + c) + (1/a + 1/b + 1/c), and q as
Rollout 2
a number with 999 nines, is divided by 1000. Hmm,
Rollout 2
etc. Wait,\n\nWait, 10^k-1 for anyk is always even? No
Rollout 2
perhaps for k ≥3, 10^k is 0 mod125?\n\nWait,
Rollout 2
anyk>=1. Then 10^k -1 is odd. Because even minus1 is
Rollout 2
2 mod 8, so 10^k mod 8 is 2^k mod
Rollout 0
a < b or a > b, so then to get the (a,b) with a < b
Rollout 5
is a segment parallel to the bases that divides the trapezoid into two regions of equal area,

DOWN_PROJ

Top 16 Positive Activations
Rollout 11
1/c = ab. So 1/a + 1/b + 1/c = bc + ac
Rollout 11
+ b + c) + (1/a + 1/b + 1/c), and q as
Rollout 11
not sure. Alternatively, p is (a + 1/a) + (b + 1/b
Rollout 14
and $q$ such that $1/p + 1/q = 1$, we have\n\n$$
Rollout 1
||2 = 2||u||2 + 2||v||2.\n\nGiven that S is
Rollout 14
to show that $I^3 \leq 8 J K$.\n\nHmm. So, the
Rollout 8
the number of possible sequences is 10 * 9 * 8 * 7 * 6
Rollout 1
1) each vector has norm 1, and 2) any two distinct vectors are orthogonal.\n\nGiven
Rollout 5
be b, then the longer base is b + 100. The midline length is (
Rollout 5
is a segment parallel to the bases that divides the trapezoid into two regions of equal area,
Rollout 0
.\n\nFirst, a rational number between 0 and 1 can be written as a fraction \(\frac
Rollout 1
that for any x, ||v_x|| = 1, and for any x ≠ x', ⟨
Rollout 4
1/C_eq = 1/C1 + 1/C2. So, I can write that
Rollout 12
{dx}{\sqrt{(x^2 - 1)(1 - \nu^2 x^
Rollout 10
\n\nWait, okay, so from 0 to 10, the losing positions (P-positions
Rollout 14
Hölder's inequality: for $p$ and $q$ such that $1/p + 1
Top 16 Negative Activations
Rollout 3
n) in m/s\n<|im_start|>assistant\n<|im_start|>think\nOkay, I need to calculate the mean molecular
Rollout 13
value to the nearest cent.\n<|im_start|>assistant\n<|im_start|>think\nOkay, so Joanie took a $6
Rollout 0
$ be the resulting product?\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I need to figure out how
Rollout 2
. Hence, answer109.\n\nTherefore, boxed answer is \boxed{109}\n\n**
Rollout 0
not a square. Therefore, the number of divisors N! So, accordingly, the number of such
Rollout 8
$m + n$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. We need to
Rollout 4
this combination of capacitors?\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to find the percentage
Rollout 2
1000$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I have this problem here:
Rollout 5
/100$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. I need to
Rollout 9
$\triangle GEM$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to solve this geometry
Rollout 1
, x2 ∈ S (with ||x1 - x2|| = d), the linear equations ⟨
Rollout 12
1 - t2 is positive (since t <1). So yes, the sqrt is real and
Rollout 15
t+1} can be written as a matrix multiplication over GF(2), where for internal lamps,
Rollout 15
be off after some time?\n<|im_start|>assistant\n<|im_start|>think\nAlright, let's try to figure out this
Rollout 0
with a unique pair (a,b) where a < b. So the count is half the number of
Rollout 11
.\n\n*David Altizio*\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I have this problem here where

Layer 20

GATE_PROJ

Top 16 Positive Activations
Rollout 10
number you add to S, you must exclude only finitely many others (i.e., n +
Rollout 8
are the only possibilities for mixed, and each is already a block of 1 red and 1 green
Rollout 0
translates as, (since 20! is fixed and a*b=20!), how many ways
Rollout 0
b?\n\nWait, 20! is a fixed number, but how is a related to b?
Rollout 7
must be a common multiple of 3 and 12, that is, a multiple of 1
Rollout 1
between centers equals d2, radii squares are each d2 / 2. So sum of radi
Rollout 8
the only possibilities for mixed, and each is already a block of 1 red and 1 green.
Rollout 15
L1 at t+1 is on. But depends on how the rules apply. Wait:\n\nThe rule
Rollout 14
2} dx is divergent at both 0 and ��. Hence this term is infinity. Thus
Rollout 10
s. Since each existing a_j rules out only finitely many (about2sqrt(a_{i
Rollout 1
are governed by\n\nFor any two spheres, there is a non-empty intersection because the distance squared between centers
Rollout 11
product is 1 modulus r^3 =1, so r=1, which is prohibited. So
Rollout 11
3 =1, so r=1, which is prohibited. So no. Hence, all three variables
Rollout 10
<=n. For each n, it's a finite number, but the set of possible moves for a
Rollout 11
as AM ≥ GM, minimum at r=1 which is 2 ), similarly if r <1,
Rollout 10
to S, you must exclude only finitely many others (i.e., n +s^2
Top 16 Negative Activations
Rollout 3
. Let's check with calculator steps.\n\nAlternatively, sqrt(28,408). Let's
Rollout 11
2/5. However, �� = nπ/3.\n\nSo if we check: for �
Rollout 5
, and then find the greatest integer not exceeding x2/100. \n\nFirst, let me
Rollout 3
7≈0.047. Thus, sqrt≈168.5 +0.0
Rollout 4
approx 2.7749, so sqrt(7.7e-5)=2.
Rollout 3
28,408.12:\n\nsqrt(28,408.12
Rollout 4
7 is 7.7e-5. sqrt(7.7e-5). Well,
Rollout 12
2))\n\nTherefore, given f(nu) = K(k), k = sqrt(1 - nu2
Rollout 11
n∈Z\n\nThus, �� = nπ/3.\n\nBut earlier, from the p condition
Rollout 3
= 28,561. So sqrt(28,436) is approximately
Rollout 5
= 18125. Then x2 /100 = 1812
Rollout 11
) e^{-iθ}\n\nBut expressions like r2 e^{i2θ} + 1/(
Rollout 3
=15.87.\n\nLinear approximation:\n\nsqrt(x + dx) �� sqrt(x) +
Rollout 4
are maximum.\n\nWait, but because (C1*C2)/(C1 + C2), if both
Rollout 3
8,408.12. So sqrt(28,408.12
Rollout 2
125. Therefore, (10^k -1) �� -1 mod12

UP_PROJ

Top 16 Positive Activations
Rollout 3
69 m/s.\n\n**Final Answer**\n\boxed{170} m/s\n\nWait no
Rollout 3
3, so if need to answer, in the absence of the temperature in the problem, I should either
Rollout 3
.5 m/s. If using calculator accurate value.\n\nSo the answer is approximately 168.
Rollout 3
69} m/s.\n\n**Final Answer**\nThe mean molecular speed of radon gas is \boxed
Rollout 1
problem statement.\n\n**[Final Answer]**\n\boxed{\text{Such a point } y \text
Rollout 4
0.88%.\n\n**Final Answer**\nThe percentage error in the calculation of the energy stored is
Rollout 14
vectors / weighet\n\n**Final Answer**\n\boxed{ \left(\int_0^\infty
Rollout 3
) in m/s\n<|im_start|>assistant\n<|im_start|>think\nOkay, I need to calculate the mean molecular speed
Rollout 9
25.\n\nPerfect.\n\n**Final Answer**\n\boxed{25}\n\n<|im_start|>answer\nSquare \(A
Rollout 11
, 3)}.\n\n**Final Answer**\n\boxed{(-1, 3)}\n\n<|im_start|>answer\n
Rollout 3
169 m/s. Thus, present answer boxed as \boxed{169 \text{
Rollout 3
So all steps are correct.\n\nThus, conclusion: at 298 K, the mean molecular speed
Rollout 3
**\nThe mean molecular speed of radon gas is \boxed{170} m/s.\n\nWait
Rollout 4
\boxed{0.88\%}.\n\n<|im_start|>answer\nTo find the percentage error in the
Rollout 0
}=128.\n\n**Final Answer**\n\boxed{128}\n\n<|im_start|>answer\nGiven a
Rollout 3
answer approximately 170 m/s. But given compute precisely, it is about 168
Top 16 Negative Activations
Rollout 2
\geq 3\), \(2^k \equiv 0 \mod 8\),
Rollout 2
=3. For k>=3:2^k mod8. But 2^3=8
Rollout 2
Thus, for k ≥3, 2^k �� 0 mod8, so (2
Rollout 2
perhaps for k ≥3, 10^k is 0 mod125?\n\nWait,
Rollout 2
k=10^{3}*10^{k-3}≡0*10^{k
Rollout 2
0^k mod 8 is 2^k mod 8.\n\nThus, each term (1
Rollout 2
is0 mod8, hence (10^k -1)= (0 -1)= -1
Rollout 2
0^k -1)≡(2^k -1). For k=1:2-
Rollout 2
maybe see if we can find when (2^k -1) �� 0 mod 8
Rollout 2
for k>=3, hence (10^k -1)≡-1 mod125
Rollout 2
k≥3: terms are 10^k -1 �� -1 mod125
Rollout 2
9s can be written as 10^n - 1. For example:\n\n- 9
Rollout 11
- ab = (r e^{iθ})2 = r2 e^{i2θ}\n-
Rollout 2
� 0 mod8, so (2^k -1) �� -1 mod8.\n\n
Rollout 7
P +56(N -P)=66N\n\nExpand:71P +56N
Rollout 14
1/q=1,\n\nab≤ (a^p)/p +(b^q)/q\n\nBut

DOWN_PROJ

Top 16 Positive Activations
Rollout 14
that would be: write $f(x) = 1 \cdot f(x)$, then $I
Rollout 11
not sure. Alternatively, p is (a + 1/a) + (b + 1/b
Rollout 1
<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to show that there exists a point y in the
Rollout 8
the number of possible sequences is 10 * 9 * 8 * 7 * 6
Rollout 15
me rephrase the rules to make sure I understand. Each lamp's next state depends only on its immediate
Rollout 12
{1}{\sqrt{(x^2 - 1)(1 - \nu^2 x^
Rollout 14
\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I have to prove this inequality: \n\n$$\n\
Rollout 1
.\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to show that there exists a point y in
Rollout 10
<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to prove that in this pebble game, there
Rollout 9
<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to solve this geometry problem where there's a square
Rollout 13
to the nearest cent.\n<|im_start|>assistant\n<|im_start|>think\nOkay, so Joanie took a $6,
Rollout 12
<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to prove that the function \( f(\nu)
Rollout 10
.\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to prove that in this pebble game,
Rollout 9
$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to solve this geometry problem where there's
Rollout 11
1/c = ab. So 1/a + 1/b + 1/c = bc + ac
Rollout 1
}$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to show that there exists a point y
Top 16 Negative Activations
Rollout 3
of some standard conditions. Hmm.\n\nWait, in kinetic theory, mean speed depends on temperature, so we
Rollout 15
lamps, the next state is sum (mod 2) of the two neighbors. Because on GF(
Rollout 4
percentage error in the calculation of the energy stored is \boxed{0.88\%}.\n\n
Rollout 3
298 (3), the least number of sig figs is 3, so result should be
Rollout 15
. However, this is equivalent to summing mod2. Because when the sum is exactly 1,
Rollout 15
cellular automata with next state being the sum mod2 of neighbor states are nilpotent if and only
Rollout 9
, the problem says the base of the triangle is EM, which in this case is the vertical side.
Rollout 10
0.\n\nThe Grundy number for n is mex{Grundy(n - s) | s is
Rollout 3
hasn't specified the temperature, this seems ambiguous. However, in many standard problems, they might take
Rollout 3
. However, in some contexts, "mean molecular speed" might refer to rms. Wait, I need
Rollout 1
d / sqrt(2), and the vectors from y to each x must be pairwise orthogonal.\n\nSo the
Rollout 11
some relationships here. Let's think about using the fact that if z is a non-real complex number,
Rollout 11
) + ab. Interesting, structuring this, maybe group terms.\n\nSimilarly, q =a/b +
Rollout 3
people use mean and average interchangeably, but in kinetic theory, they are different. Let me make sure
Rollout 15
as defined (every cell updates to the sum mod2 of its neighbors, ends have one neighbor). The
Rollout 12
ularity. So perhaps the Leibniz formula might still apply, but the term involving the upper limit

Layer 21

GATE_PROJ

Top 16 Positive Activations
Rollout 0
then the number of coprime pairs with a < b is exactly half of the total number of cop
Rollout 0
factors, number of coprime pairs with a < b and a*b=20! is
Rollout 15
if and only if exactly one of its neighbors is on at t. So if a lamp at t has
Rollout 6
S_i has at most p(n) strings of length n, total is k*p(n). Since k
Rollout 10
position. Wait, but n=0 is a losing position, as the player to move loses immediately.
Rollout 6
_i has at most p(n) strings of length n, total is k*p(n). Since k is
Rollout 1
of closed sets with the finite intersection property is non-empty if the space is compact. However, Hilbert
Rollout 10
from n=5, can we move to a losing position?\n\n5-1=4 (winning
Rollout 0
factors. The number of coprime pairs a < b where a*b=720 is
Rollout 10
the second player. So n=2 is a losing position, so P-position.\n\nWait, contradicts
Rollout 14
�0^∞ x f(λ x) dx = λ^{a -2} ��0
Rollout 14
term: ||f(x) x^{1/2} ||_2=(∫f(x)^2
Rollout 10
positions (next player can force a win) or P-positions (previous player can force a win).
Rollout 10
, is the player to move next on n=1 the second player? Wait, here's the initial
Rollout 14
\left(\int_0^\infty xf(x) dx \right) $$ \n<|im_start|>assistant\n
Rollout 14
\left(\int_0^\infty xf(x) \, dx \right) }\n\]
Top 16 Negative Activations
Rollout 5
but these are not similar necessarily. \n\nAlternatively, think algebraically. Let me denote the original trape
Rollout 8
are adjacent and all greens are adjacent.\n\nAlternatively, think of the problem as the laid-out 5 cards
Rollout 2
of1000 here?\n\nAlternatively, maybe because many of the terms in the product from k=
Rollout 1
system.\n\nBut how does this help?\n\nAlternatively, think of y as a solution to the system of equations
Rollout 3
\boxed{169}.\n\nBut initially I thought 168.5 rounded to
Rollout 10
every other pair. Let me think. Alternatively, maybe in each two steps, the losing positions are even
Rollout 10
.\n\nBut given that my earlier attempt at periodicity was contradicted at n=25, the Grund
Rollout 4
figures, 0.88%.\n\nAlternatively, maybe give it as 0.877%,
Rollout 5
step. Let me verify quickly. \n\nAlternatively, perhaps it is quicker to recall the formula for the line
Rollout 15
3 over GF(2). Maybe my error was in the initial simulation.\n\nFirst, for n=3
Rollout 2
multiple of1000 here?\n\nAlternatively, maybe because many of the terms in the product from k
Rollout 9
. Since each side is 10 units. So AI would be vertical, IM horizontal, ME vertical
Rollout 11
computation.\n\nRecall that from abc =1, we can perhaps normalize variables, such that leaving to substitute
Rollout 11
no immediate simplification. However, if I consider that abc = 1, we can make substitutions.
Rollout 8
color sequences.\n\nTherefore, if color sequences are counted based on color orders, there are 10 happy
Rollout 7
. But how does this make sense?\n\nAlternatively, perhaps during the problem, the composition of promoted and repe

UP_PROJ

Top 16 Positive Activations
Rollout 14
1/4} + ( ��J )) /2.\n\n Well, no, the expression is
Rollout 6
the machine M, given x of length n, needs to get the bits of α_n, which is
Rollout 7
total promoted score =79*(P +x).\n\nBut original promoted now scores higher. New scores=
Rollout 7
+x)=79*(16 +2)=79*18=1422
Rollout 13
compared to if it compounds annually. Hmm, I need to calculate the amount owed in both cases and then
Rollout 14
0^AA f2 dx.\n\nSimilarly, for I2= ��A^infty f dx
Rollout 12
) > (1 -k2) K(k). Let me check for some value. For k approaching
Rollout 7
-promoted total=47*(R -x). Then:\n\n79(P +x) +4
Rollout 13
in both cases and then find the difference between them. Let me think step by step.\n\nFirst, let
Rollout 4
2*0.004 )^2 )\n\nCompute:\n\n0.003605
Rollout 14
:\n\nLeft-hand side: (a b)^3,\n\nRight-hand side:8 * (a^2
Rollout 1
+ ||x'||2 - d2)/2 - ⟨x + x', y⟩ + ||y
Rollout 1
since k depends on ||y||2, our goal is to find y and k such that this equation
Rollout 3
793 * 0.222 ��0.697433.\n\n
Rollout 11
1 ) + (1/t1 +t1 ) + (1 +1)= 2(t1
Rollout 9
square is 80 square units, and I need to find the length of the altitude to EM in
Top 16 Negative Activations
Rollout 12
K(k) = [ nu / (1 - nu2 ) ] K(k)\n\nSo altogether,\n\nf
Rollout 1
bert space, even a countable set can be dense.\n\nAlternatively, we can use a mapping where y
Rollout 1
of arbitrary cardinality, but perhaps construction requires transfinite induction.\n\nAlternatively, if we realize that the set
Rollout 2
3, and 10 is 2*5, so 10 and 125
Rollout 12
) nu ) + [ nu / (1 - nu2) ) ] K(k)\n\n= [ nu
Rollout 1
are weakly compact.\n\nBut given that spheres are closed and convex. Wait, the sphere { y :
Rollout 1
since H is infinite-dimensional, it is a complete metric space. However, the intersection might still be empty
Rollout 1
right. The sphere is not convex. So perhaps closed balls would be convex, but spheres aren't.
Rollout 1
averaging may not work. But perhaps we can use completeness of the Hilbert space.\n\nAlternatively, if H
Rollout 2
(3) * (-1)^{997} mod8.\n\nFirst, compute (-1)^
Rollout 8
number is 10P5 = 10! / (10-5)! =
Rollout 1
is non-empty. So, since each sphere is closed and convex, and in a Hilbert space,
Rollout 1
ogonality condition comes automatically from the equidistant condition. So that's interesting. Therefore, if
Rollout 14
.\n\nAlternatively, maybe use Hölder’s inequality. Holder's for triples: 1/p +1/q
Rollout 11
Similarly, p = x + y + z + xy + yz + zx\n\nBut xyz =
Rollout 0
that \(a\) and \(b\) are coprime factors of 20!. So they form

DOWN_PROJ

Top 16 Positive Activations
Rollout 11
b + c) + (ab + bc + ca) + 2. Wait, so then p
Rollout 9
square, then the sides are AI, IM, ME, and EA. Wait, maybe that's a
Rollout 5
area. \n\nFor a line segment parallel to the bases that divides the trapezoid into two regions
Rollout 11
b +c) + (ab + bc + ca)= (a +b +c) + (
Rollout 11
1/(xy) + xy + y/(xy) + x/(xy) = x + y
Rollout 11
b + c) + (ab + bc + ca). So p = a + b + c +
Rollout 2
Hmm, okay, let's unpack this step by step.\n\nFirst, the problem is asking for the remainder
Rollout 15
4 (all lamps on), the system actually reaches all off in 5 steps. However, with a
Rollout 3
for approximation.\n\nWait, given that no temperature is given, perhaps they just want the formula, but I
Rollout 11
b + c) + (ab + bc + ca). So that's correct. So p = (
Rollout 11
b +c) + (ab + bc + ca ) +2.\n\nSo p= s + t
Rollout 9
, which is 10; the bases are parallel sides? But wait the two lines at x=
Rollout 11
b + c) + (ab + bc + ca) + 1 = (a + b +
Rollout 11
b + c) + (ab + bc + ca).\n\nSo that's equivalent to (1 + a
Rollout 11
b +c) + (ab + bc + ca) + abc=1. But abc=1
Rollout 11
a + b + c + ab + bc + ca. That might be easier to work with. So
Top 16 Negative Activations
Rollout 1
such sets have certain properties.\n\nAlternatively, I might need to use the parallelogram law. Let me
Rollout 15
off after some time?\n<|im_start|>assistant\n<|im_start|>think\nAlright, let's try to figure out this problem
Rollout 6
each encoding a bit of α_n. However, then S_L would have p(n) strings for each
Rollout 11
*David Altizio*\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I have this problem here where a
Rollout 3
mathematics assistant.\n<|im_start|>user\nCompute the mean molecular speed v in the heavy gas radon (Rn
Rollout 10
how his opponent plays.\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to prove that in this
Rollout 2
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nFind the remainder when
Rollout 13
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nJoanie takes a
Rollout 3
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nCompute the mean molecular
Rollout 6
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet's say a
Rollout 4
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nTwo capacitors with
Rollout 7
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nIn a mathematics test
Rollout 8
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nKathy has $
Rollout 10
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nConsider the following two
Rollout 9
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nSquare $AIME
Rollout 11
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nSuppose $

Layer 22

GATE_PROJ

Top 16 Positive Activations
Rollout 1
$ \mathcal{H}$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to show
Rollout 0
**\n\boxed{128}\n\n<|im_start|>answer\nGiven a rational number between 0 and 1
Rollout 2
109 in both cases. So The answer is109. Huh.\n\nBut wait,
Rollout 13
{187.12}\n\n<|im_start|>answer\nTo determine how much more money Joanie will owe
Rollout 13
quarterly compared to if it compounds annually. Hmm, I need to calculate the amount owed in both cases and
Rollout 6
a sparse oracle, which is standard.\n\n<|im_start|>answer\n**Problem 1:**\n\nTo construct the sparse
Rollout 10
win no matter how his opponent plays.\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to prove
Rollout 15
lamps will be off after some time?\n<|im_start|>assistant\n<|im_start|>think\nAlright, let's try to figure
Rollout 4
0.88\%}.\n\n<|im_start|>answer\nTo find the percentage error in the calculation of the
Rollout 0
128 such numbers. Therefore, The answer is 128, corresponding to \boxed{
Rollout 9
EM. Let me try to visualize this first. \n\nLet me draw the square AIME. Since it
Rollout 15
\) being powers of two. Hence the answer is \boxed{2^k} for integers \(
Rollout 10
second player can force a win.}\n\n<|im_start|>answer\nTo prove that there are infinitely many initial situations in
Rollout 11
,q)$ .\n\n*David Altizio*\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I have this problem
Rollout 6
$L$ . [/*]\n<|im_start|>assistant\n<|im_start|>think\nOkay, let me try to work
Rollout 0
is 128. Hence, the answer is 128?\n\nBut to confirm this,
Top 16 Negative Activations
Rollout 3
811.04. Divide by π*0.222≈0.69
Rollout 3
.314*298/(π*0.028)) �� sqrt(
Rollout 3
.314*298/(π*0.222)) �� sqrt(
Rollout 9
: y= (-5/25)x +50/25= (-1/5)x
Rollout 3
as earlier for radon)\n\nDenominator: pi*0.028= ~0.0
Rollout 3
,953.6; divided by π*0.222 �� 0.
Rollout 3
3 K:\n\n66.512 *293= compute 66.51
Rollout 10
7...\n\nSo starting at 0:\n\n0 +2=2\n\n2 +3=5\n\n5
Rollout 7
new average 59. 54 +5=59. Correct.\n\nSo N=1
Rollout 3
19,811.04 /0.697433≈28
Rollout 4
Multiply by Δa:0.36 *10= 3.6 pF\n\nSecond
Rollout 7
76, as their original was71 +5) and x repeaters with average A+5
Rollout 3
in calculation:\n\n8 * R * T/(π*M): all three terms are three sig figs minimum
Rollout 3
0.576\n\nDenominator: π * M = π * 0.222
Rollout 4
energy:\n\nE_nominal=0.5 *1200e-12 *25
Rollout 14
r=4, since 1/2 +1/4 +1/4=1.\n\nThen

UP_PROJ

Top 16 Positive Activations
Rollout 4
1/C_eq = 1/C1 + 1/C2. So, I can write that as
Rollout 14
, then q=for 1/p +1/q=1. So q=2. But again
Rollout 10
Then for n >=1:\n\nGrundy[n]= mex{Grundy[n - s] | s is
Rollout 6
sum over i of |S_i^{=n}|, which is at most k*p(n). Since
Rollout 4
get maximum energy, since energy is (1/2)*C_eq*V2, maximum energy would
Rollout 4
energy. The energy stored is (1/2) * C_eq * V^2. So,
Rollout 6
1 to k of |S_i^{=n}|. Since each S_i has at most p(n
Rollout 10
5.\n - The Grundy numbers (mex values) were computed for several positions, showing non
Rollout 11
the function f(r) =r3 +1/r3.\n\nIf r >1, then f(r
Rollout 11
and b.\n\n p= a + b +1/(ab) +ab + a/(ab) +
Rollout 11
In上述 p: p= t1 +1/t1 + t2 +1/t2 + t
Rollout 11
a + b + c + ab + bc + ca} = \overline{a} + \
Rollout 7
, total score becomes 66N + 5N = 71N. But also,
Rollout 12
sqrt( (t2 - ν2)(1 - t2) ), which for t in (ν
Rollout 9
, and their common area is 80 square units. Since the square itself is 10x
Rollout 4
in a capacitor is (1/2) * C_eq * V2. Since the question is about
Top 16 Negative Activations
Rollout 2
is \boxed{109}\n\n**Final Answer**\n\boxed{109}\n\n<|im_start|>answer
Rollout 4
is approximately 0.88%.\n\n**Final Answer**\nThe percentage error in the calculation of the energy
Rollout 3
{169} m/s.\n\n**Final Answer**\nThe mean molecular speed of radon gas is
Rollout 13
$187.12.\n\n**Final Answer**\n\boxed{187.12
Rollout 5
81 is the correct answer. \n\n**Final Answer**\n\boxed{181}\n\n<|im_start|>answer
Rollout 9
being h=25.\n\nPerfect.\n\n**Final Answer**\n\boxed{25}\n\n<|im_start|>answer\n
Rollout 15
neighbor L2 is on. L2 remains on since one neighbor on (L3). Wait, but
Rollout 8
26= 157.\n\n**Final Answer**\n\boxed{157}\n\n<|im_start|>answer
Rollout 11
?\n\nBut let's assume for simplicity that all variables are real? Wait, but the problem states that none
Rollout 7
, part(b) has no solutions.\n\n**Final Answer**\n\n(a) \boxed{12}, \
Rollout 13
187.12}\n\n<|im_start|>answer\nTo determine how much more money Joanie will owe if
Rollout 8
6= 157.\n\n**Final Answer**\n\boxed{157}\n\n<|im_start|>answer\n
Rollout 5
Wait, let me check the calculations again step by step. Probably, I made an algebraic mistake.\n\n
Rollout 11
{(-1, 3)}.\n\n**Final Answer**\n\boxed{(-1, 3)}\n\n
Rollout 2
\boxed{109}\n\n**Final Answer**\n\boxed{109}\n\n<|im_start|>answer\n
Rollout 3
{169} m/s.\n\n**Final Answer**\nThe mean molecular speed of radon gas is

DOWN_PROJ

Top 16 Positive Activations
Rollout 7
71N.\n\nWhich simplifies as we did before to x=N/12.\n\nHowever, after
Rollout 7
-promoted $47$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I'm trying to
Rollout 10
win no matter how his opponent plays.\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to prove
Rollout 5
the two bases. Wait, I might need to derive that. \n\nSuppose the original trapez
Rollout 2
since 1000 is 8 * 125 and 8 and 12
Rollout 7
)=71N.\n\nWhich simplifies as we did before to x=N/12.\n\nHowever,
Rollout 15
lamps will be off after some time?\n<|im_start|>assistant\n<|im_start|>think\nAlright, let's try to figure
Rollout 15
3: Three lamps in a line. Let's label them L1, L2, L3.
Rollout 14
fty $$ Prove the following inequality. $$ \left(\int_0^\infty f
Rollout 11
= ac, 1/c = ab, as before. So 1/a is bc. Let me
Rollout 14
they are not; the inequality must hold for all valid f), this equation doesn’t hold in general.
Rollout 14
(x) dx \right) $$ \n<|im_start|>assistant\n<|im_start|>think\nOkay, so I have to prove
Rollout 11
approach, start by assuming that p and q take minimal values, or algebraic manipulation.\n\nFrom p =
Rollout 12
[ 1 / sqrt(1 - k2 cos2theta) ] dtheta\n\nBut 1
Rollout 13
to the nearest cent.\n<|im_start|>assistant\n<|im_start|>think\nOkay, so Joanie took a $6,
Rollout 10
positions are infinite.\n\nPerhaps applying the concept from Wythoff's game, but with squares.\n\nWait,
Top 16 Negative Activations
Rollout 8
the number of sequences is the number of injective functions from 5 positions to 10 cards,
Rollout 4
/C_eq = 1/C1 + 1/C2. So, I can write that as C
Rollout 4
1/C_eq = 1/C1 + 1/C2. So, I can write that as
Rollout 2
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nFind the remainder when
Rollout 13
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nJoanie takes a
Rollout 0
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nGiven a rational number
Rollout 3
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nCompute the mean molecular
Rollout 1
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet $ \
Rollout 6
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet's say a
Rollout 11
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nSuppose $
Rollout 7
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nIn a mathematics test
Rollout 4
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nTwo capacitors with
Rollout 8
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nKathy has $
Rollout 10
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nConsider the following two
Rollout 9
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nSquare $AIME
Rollout 12
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nProve that the

Layer 23

GATE_PROJ

Top 16 Positive Activations
Rollout 4
in C_eq: sqrt( ( (10*3000^2)/(200
Rollout 4
5 -10*3000 +10*15=6,000
Rollout 4
\right)^2 + \left( 2 \cdot \frac{\Delta V}{V} \
Rollout 11
+ 1/(a2 z ) + a*(a z ) + (a z )*1/(
Rollout 12
)\n\n= (1 - ν2 - t2 + ν2)/(1 - ν2) ) =
Rollout 3
14 * 298 / (π * 0.222)) Hmm, okay
Rollout 7
x +59*(N/3) -59x =71N\n\nCalculate the terms
Rollout 4
03605)^2 + (2*0.004 )^2 )\n\nCompute
Rollout 7
:\n\n75*(2N/3) +75x +59*(N/3)
Rollout 11
a z )*1/(a2 z ) +1/(a2 z )* a\n\nSimplify term
Rollout 12
/dθ = [ (1 - ν2) * 2 sinθ cosθ ] / [
Rollout 4
/(5000)^2= (4*10^6)/(25*10
Rollout 11
r3*r +r3*1/r +1/r3*r +1/r3*1/r
Rollout 11
2 z ) + a*(a z ) + (a z )*1/(a2 z ) +
Rollout 13
, check using the formula difference.\n\nDifference = P*( (1 + r/4)^(4t
Rollout 11
1/(a*b ) =1/(a^2 z). Hence, c =1/(a^
Top 16 Negative Activations
Rollout 2
2^3 = 8 �� 0 mod 8, and 2^k for
Rollout 15
]], which is identity. Therefore, not nilpotent. For n=3, as seen earlier,
Rollout 10
7-16=1, which is a winning position for opponent. So all moves from n=
Rollout 10
n=25 has Grundy number3≠0, so it's a winning position. Hence,
Rollout 10
Grundy[2} = mex{1}=0.\n\nn=3: subtract1→2,
Rollout 2
with remainders, especially modulo 1000, it often helps to look at the problem modulo
Rollout 2
0. But891 mod1000* (-1)= -891 mod1
Rollout 2
2^k for k ≥3 is 0 mod8. Thus, for k ≥3,
Rollout 10
a losing position. So n=4 is a winning position.\n\nWait, that would mean n=4
Rollout 10
14-4=10 is a losing position. So n=14 is winning.\n\n
Rollout 10
5 (losing). So n=16 is winning.\n\nn=17: subtract 1=
Rollout 2
\n\n2^3 = 0 mod 8\n\n2^4 = 0 mod 8\n\n
Rollout 10
2 (losing), so n=13 is winning.\n\nn=14: subtract 1=
Rollout 10
ning) or 4-4=0 (losing). So if you take 4, leaving
Rollout 10
(losing). Therefore, n=21 is winning.\n\nn=22: Check if it's
Rollout 11
≠ 0 or π, so sinθ ≠0. Hence, 2 cosθ = r3

UP_PROJ

Top 16 Positive Activations
Rollout 5
regions of equal area. \n\nFor a line segment parallel to the bases that divides the trapezoid
Rollout 12
ν of dx over sqrt( (x2 -1)(1 - ν2x2) ). Let
Rollout 1
x - y) x ∈ S, forms an orthonormal system. This means:\n\n• For each
Rollout 3
.\n\n66.512 * 273 �� 66.512
Rollout 14
I ≤ ||f(x)x^(-1/3)||_a *||x^(1/3
Rollout 14
frac{f(x)}{x^{1/3}} \cdot x^{1/3} dx
Rollout 1
Zorn's lemma, to find a maximal orthonormal system. But howdoes that help.\n\nHmm
Rollout 2
. For k>=3:2^k mod8. But 2^3=8≡0
Rollout 12
sqrt( ( (t/ν)^2 -1 )(1 - t^2) )\n\nHmm.
Rollout 2
:\n\n9*99=891 mod1000\n\n891*99
Rollout 2
:\n\n(5m +5)≡5 mod8\n\nSubtract 5:\n\n5m≡0
Rollout 12
{dt}{\sqrt{(t^2 - \nu^2)(1 - t^2)}
Rollout 1
x ∈ S, forms an orthonormal system. This means:\n\n• For each x ∈ S,
Rollout 1
radius d / sqrt(2) centered at each point.\n\nBut in R^3, let's compute
Rollout 2
^k -1) �� -1 mod8.\n\nTherefore, for k=1: (2
Rollout 3
the mean (average) speed derived from the Maxwell-Boltzmann distribution:\n\n\[\nv = \
Top 16 Negative Activations
Rollout 8
26 are coprime, so yeah, simplified.\n\nTherefore, the probability is 31/
Rollout 0
and 1. When written in lowest terms, numerator a and denominator b must satisfy that. However,
Rollout 7
56*8=448. Total of promoted-from-repeaters(original avg A)x=2
Rollout 11
1, we can perhaps normalize variables, such that leaving to substitute variables. For example, let me set
Rollout 10
because you can take 1 and leave the opponent with 0. n=2 is a losing position
Rollout 5
the ratio of the bases is the same as the ratio of the heights. Wait, but these are not
Rollout 7
1$ and that of the repeaters $56$ . \nHowever, due to an
Rollout 3
inputs: 8 (exact)( since formula has 8, which is exact), R=8.
Rollout 6
related to n, but over all lengths, the total number for each length would be polynomial.\n\nWait,
Rollout 7
: their total original=124, new total=124 +5*2=1
Rollout 8
(5,5−k).\n\nThen we totaled 1200 +2400 +
Rollout 13
Then, since both numbers have two decimal places, total decimals places would be 4, so 1
Rollout 0
\(a \neq b\). Hence, total coprime pairs are \(2^{8}\
Rollout 7
): total score original=186, new total=186 +5*3=2
Rollout 7
0-64, sum /x=A must be possible?\n\nLet x=N/12. For
Rollout 5
+ b)/2 * h]. Then, simplifying, (a + x)k = (a

DOWN_PROJ

Top 16 Positive Activations
Rollout 0
k is primes. But, since prime factors with exponents: Wait, but each prime is either included
Rollout 0
factorization of 20!.\n\nTo compute the exponents of primes in 20!:\n\nFor
Rollout 0
not a square (unless for example primes with even exponents), it depends. As with N! for
Rollout 0
that's not correct. Each must have all primes exponents from 20! For example, if
Rollout 0
For each prime p | N, assign all the exponents in p to a or to b, but
Rollout 0
, actually 20! has prime factors with exponents. For example, 2^18
Rollout 0
is not a square (due to the primes with exponents as discussed earlier), there are no pairs where
Rollout 0
a = product_{p in S} p^{e_p}, and b = 20! /
Rollout 5
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nOne base of a
Rollout 0
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nGiven a rational number
Rollout 1
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet $ \
Rollout 4
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nTwo capacitors with
Rollout 6
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet's say a
Rollout 3
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nCompute the mean molecular
Rollout 8
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nKathy has $
Rollout 7
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nIn a mathematics test
Top 16 Negative Activations
Rollout 12
P. Turan]\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to prove that the function
Rollout 11
*David Altizio*\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I have this problem here where a
Rollout 14
\right) $$ \n<|im_start|>assistant\n<|im_start|>think\nOkay, so I have to prove this inequality:
Rollout 1
cal{H}$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to show that there exists
Rollout 10
how his opponent plays.\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to prove that in this
Rollout 11
.\n\n*David Altizio*\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I have this problem here where
Rollout 15
off after some time?\n<|im_start|>assistant\n<|im_start|>think\nAlright, let's try to figure out this problem
Rollout 14
dx \right) $$ \n<|im_start|>assistant\n<|im_start|>think\nOkay, so I have to prove this inequality
Rollout 7
$47$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I'm trying to solve this math
Rollout 15
be off after some time?\n<|im_start|>assistant\n<|im_start|>think\nAlright, let's try to figure out this
Rollout 6
. [/*]\n<|im_start|>assistant\n<|im_start|>think\nOkay, let me try to work through these problems
Rollout 8
m + n$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. We need to find
Rollout 9
triangle GEM$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to solve this geometry problem
Rollout 0
$ be the resulting product?\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I need to figure out how
Rollout 12
[P. Turan]\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to prove that the
Rollout 1
mathcal{H}$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to show that there

Layer 24

GATE_PROJ

Top 16 Positive Activations
Rollout 3
Pressure), which is 0°C and 1 atm. However, sometimes people use 25°C
Rollout 2
\n\nk=2:10^2 -1=99 mod125\n\nk=
Rollout 0
20/8) + floor(20/16) =10+5+2
Rollout 2
\n\nk=4:10^4 -1=9999≡(10^
Rollout 0
20/2) + floor(20/4) + floor(20/8)
Rollout 3
STP is 1 bar instead of 1 atm, but regardless, the formula for speed only depends
Rollout 3
73.15 K) and 1 atm. In physics, sometimes it's 20
Rollout 2
\n\nk=3:10^3 -1=999≡-1 mod12
Rollout 2
;\n\nk=2,10^2 -1=99\n\nk=3, so1
Rollout 3
8 K.\n\nCompute v_avg = sqrt(8*R*T/(π*M))\n\nLet's calculate the numerator
Rollout 0
20/4) + floor(20/8) + floor(20/16
Rollout 0
\n\nFor prime 3: floor(20/3) + floor(20/9)
Rollout 0
20/3) + floor(20/9) + floor(20/27
Rollout 10
. So Grundy[2} = mex{1}=0.\n\nn=3: subtract1→
Rollout 3
273.15 K and 1 atm. At STP, the molar volume is
Rollout 0
20/9) + floor(20/27)=6 +2 +0=8
Top 16 Negative Activations
Rollout 11
is 1. \n\nFirst, let me note that since abc = 1, maybe we can write
Rollout 11
Maybe considering algebra instead of analysis. Suppose we note that given p and q are real, can we relate
Rollout 5
oid into equal areas. I think the formula is that the length of this line is the root mean square
Rollout 11
me note that since abc = 1, maybe we can write 1/a = bc, 1
Rollout 3
maybe the term "mean molecular speed" is defined such that it doesn't require temperature because it's given
Rollout 11
symmetry.\n\nBut maybe for this to hold, require that a/b = conjugate of y/z, since
Rollout 11
1, we can make substitutions. Let's note that since abc = 1, any term can be
Rollout 5
two bases. Wait, I might need to derive that. \n\nSuppose the original trapezoid
Rollout 11
p and q through symmetric expressions.\n\nAlternatively, notice that q is related to some expression of p. For
Rollout 3
(3RT/M). So perhaps the problem wants the average speed. However, in some contexts, "
Rollout 5
100. \n\nFirst, let me recall some trapezoid properties. In a trape
Rollout 3
K.\n\nAlternatively, maybe they expect us to state that the temperature is required? But no, the problem
Rollout 3
, maybe the term "mean molecular speed" is defined such that it doesn't require temperature because it's
Rollout 11
pairs (p, q).\n\nMaybe using the fact that if a complex number is real, its complex conjug
Rollout 11
.\n\nAlternatively, since abc =1, perhaps consider that one of the variables is the product of conjugates
Rollout 11
must have non-real coefficients. Hence, perhaps looking for cases where p and q have specific relationships leading to

UP_PROJ

Top 16 Positive Activations
Rollout 2
2^k -1) �� 0 mod 8, which would make the entire product congr
Rollout 9
the triangle GEM would be a large triangle with base EM (the right edge of the square) and
Rollout 2
3, 2^k �� 0 mod8, so (2^k -1)
Rollout 0
b\).\n\nNow, how do I count such pairs?\n\nAlternatively, in number theory, this is equivalent
Rollout 9
has two equal sides (GE and GM) with base EM.\n\nSo let's calculate coordinates.\n\nGiven:\n\n
Rollout 0
number of coprime pairs \((a, b)\) where \(a \times b =
Rollout 9
5) where h>10. Then the base EM is vertical from (10,0)
Rollout 2
≥3, 10^k≡0 mod125. Therefore, (10^
Rollout 0
number of coprime pairs \((a, b)\) with \(a < b\) is \
Rollout 2
} -1). We need to compute P mod 1000.\n\nNow, let's approach
Rollout 9
The area of this triangle can be calculated. The base EM is same as base from (10,
Rollout 2
x in:\n\nx �� 109 mod125\n\nx ��5 mod8
Rollout 2
\[\n x \equiv 5 \mod 8 \quad \text{and} \
Rollout 2
9=891. 891 mod125=16, then 1
Rollout 0
is equivalent to: how many pairs (a, b) are there with gcd(a, b)=1
Rollout 2
2^k for k ≥3 is 0 mod8. Thus, for k ≥3,
Top 16 Negative Activations
Rollout 14
which contradicts the given that $f(x) \geq 0$. Thus, perhaps f(x
Rollout 12
'(\nu) - F(a(\nu), \nu) \cdot a'(\nu) +
Rollout 14
inequality holds for exponential functions, but they don't reach the equality. This suggests that perhaps the optimal function
Rollout 14
int_0^\infty xf(x) dx <\infty $$ Prove the following inequality
Rollout 14
1 ≤4. True, but again doesn't reach equality. So again, the equality is not achieved
Rollout 5
2 = 75 ± 5*sqrt(725)\n\nSince y = x +
Rollout 3
298 K), their average speeds can be in the hundreds of m/s. For example, for
Rollout 5
(18125) which is exactly what we have. Thus, that is consistent.\n\nTherefore
Rollout 12
ularity that is integrable, the differentiation can still go through as long as the integral of the derivative conver
Rollout 5
* 725) = 5 * sqrt(725). Is that right?\n\nWait
Rollout 4
1 * C2) / (C1 + C2). Then, the energy stored in a capacitor
Rollout 1
any two vectors from the centroid is a constant negative value. Wait, perhaps scaling them appropriately could make them
Rollout 3
no, temperature must be in Kelvin in the formulae.\n\nHence, given the confusion, perhaps the
Rollout 7
, due to an error in the wording of a question, all scores are increased by $5$
Rollout 3
compute the speed? Maybe they just want the formula expressed in terms of molar mass and R? No
Rollout 14
<1, so inequality holds but again doesn't achieve equality.\n\nSo maybe equality is not achievable for these

DOWN_PROJ

Top 16 Positive Activations
Rollout 2
0 mod8\n\nThus, 5m≡0 mod8. So 5m is divisible by
Rollout 7
, but reading the problem again: "due to an error in the wording of a question, all scores
Rollout 0
\) are coprime factors of 20!. So they form a coprime pair whose product
Rollout 3
required? But no, the problem says "compute the mean molecular speed," so they must want a numerical
Rollout 11
.\n\n*David Altizio*\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I have this problem here where
Rollout 12
[P. Turan]\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to prove that the
Rollout 2
8\n\nSubtract 5:\n\n5m≡0 mod8\n\nThus, 5m≡0
Rollout 0
are related to the divisors of 20! where each pair consists of two coprime numbers
Rollout 4
this combination of capacitors?\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to find the percentage
Rollout 6
$ . [/*]\n<|im_start|>assistant\n<|im_start|>think\nOkay, let me try to work through these
Rollout 9
, the apex is G. And the altitude to EM is the perpendicular distance from G to EM. Since
Rollout 7
's read the original problem again:\n\n"due to an error in the wording of a question, all scores
Rollout 5
/100$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. I need to
Rollout 3
n) in m/s\n<|im_start|>assistant\n<|im_start|>think\nOkay, I need to calculate the mean molecular
Rollout 1
mathcal{H}$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to show that there
Rollout 10
matter how his opponent plays.\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to prove that in
Top 16 Negative Activations
Rollout 8
they can be simplified. Since 31 is prime. 126 divides by 2,
Rollout 14
2} dx is divergent at both 0 and ��. Hence this term is infinity. Thus
Rollout 10
a winning position. Wait, but 25≡0 mod5.\n\nWait, that contradicts the
Rollout 14
^{-2} dx is divergent at both 0 and ��. Hence this term is infinity.
Rollout 11
, r3 becomes large, so function tends to infinity. The minimum of f(r) occurs at r
Rollout 0
and \(b\) are coprime, each prime factor in 20! must go entirely to
Rollout 5
x + 9375\n\nSubtract 125x + 9375
Rollout 9
to the start? Wait, now I'm confused. Wait squares have four sides. Let me actually assign
Rollout 9
+ y^2\n\nSubtract y^2 from both sides:\n\n0 = 100 -
Rollout 0
256\), but each pair is counted twice except when \(a = b\). But since
Rollout 5
2b + 150; subtract 2b from both sides: b + 7
Rollout 9
that would be the vertical side. Wait, no, in that labeling, if AIME is the square
Rollout 5
5x + 9375\n\nSubtract 125x + 937
Rollout 5
0 = 125y\n\nBring all terms to left side:\n\ny^2 - 2
Rollout 5
+ 150; subtract 2b from both sides: b + 75 =
Rollout 9
point G which is inside the square. Wait no, if the altitude is from G to EM, which

Layer 25

GATE_PROJ

Top 16 Positive Activations
Rollout 3
(3RT/M), where R is the gas constant, T is the temperature in Kelvin, and M
Rollout 10
1, which is winning. Hmm, wait. So if from n=5, can we move
Rollout 10
and odd. But maybe every other pair. Let me think. Alternatively, maybe in each two steps,
Rollout 15
O F O F (t=0). Let me step through that.\n\nt=0: O F
Rollout 15
transition might turn OO into something else? Let me check:\n\nWait, if both lamps are on at t
Rollout 10
3. Is there a pattern? Let me check n=11.\n\nn=11:
Rollout 3
8 K.\n\nCompute v_avg = sqrt(8*R*T/(π*M))\n\nLet's calculate the numerator
Rollout 3
the formula.\n\nv_avg = sqrt((8 * R * T) / (π * M))\n\nR
Rollout 10
move leaves the opponent in a winning position. Let's verify n=3.\n\nn=3: possible
Rollout 10
, 3. Is there a pattern? Let me check n=11.\n\nn=11
Rollout 15
(2) would have different behavior. Let me check.\n\nWait, okay, let's model the transitions
Rollout 10
, larger square numbers alter the periodicity. Let me recast the previous positions and see the exact classifications
Rollout 15
of two, the system converges. Let me check n=4 (which 2^2).
Rollout 10
mean n=4 is winning.\n\nWait a moment. Let me write them down:\n\n0: losing\n\n
Rollout 15
a linear algebra problem over GF(2). Let me think.\n\nGiven that the next state is determined by
Rollout 15
another transition might turn OO into something else? Let me check:\n\nWait, if both lamps are on at
Top 16 Negative Activations
Rollout 2
\boxed{109}\n\n**Final Answer**\n\boxed{109}\n\n<|im_start|>answer\n
Rollout 11
(-1, 3)}.\n\n**Final Answer**\n\boxed{(-1, 3)}\n\n<|im_start|>
Rollout 3
169} m/s.\n\n**Final Answer**\nThe mean molecular speed of radon gas is \
Rollout 3
m/s, and rounded to three sig figs is 169 m/s. Thus, present
Rollout 0
-1}=128.\n\n**Final Answer**\n\boxed{128}\n\n<|im_start|>answer\n
Rollout 9
h=25.\n\nPerfect.\n\n**Final Answer**\n\boxed{25}\n\n<|im_start|>answer\nSquare
Rollout 1
as per problem statement.\n\n**[Final Answer]**\n\boxed{\text{Such a point } y
Rollout 7
no solution.\n\nBut the problem is translated probably expecting answer. Maybe my initial steps overlooked something.\n\nWait,
Rollout 8
6= 157.\n\n**Final Answer**\n\boxed{157}\n\n<|im_start|>answer\n
Rollout 0
}=128.\n\n**Final Answer**\n\boxed{128}\n\n<|im_start|>answer\nGiven a
Rollout 4
approximately 0.88%.\n\n**Final Answer**\nThe percentage error in the calculation of the energy stored
Rollout 5
1 is the correct answer. \n\n**Final Answer**\n\boxed{181}\n\n<|im_start|>answer\n
Rollout 13
187.12.\n\n**Final Answer**\n\boxed{187.12}\n\n
Rollout 13
7.12.\n\n**Final Answer**\n\boxed{187.12}\n\n<|im_start|>answer
Rollout 5
the correct answer. \n\n**Final Answer**\n\boxed{181}\n\n<|im_start|>answer\nLet the
Rollout 9
25.\n\nPerfect.\n\n**Final Answer**\n\boxed{25}\n\n<|im_start|>answer\nSquare \(A

UP_PROJ

Top 16 Positive Activations
Rollout 11
4. But c = -1/4 is real, which is prohibited. Not allowed. So,
Rollout 14
vectors / weighet\n\n**Final Answer**\n\boxed{ \left(\int_0^\infty
Rollout 10
win in these cases.\n\n**Final Answer**\n\boxed{There are infinitely many initial situations where the second
Rollout 9
25.\n\nPerfect.\n\n**Final Answer**\n\boxed{25}\n\n<|im_start|>answer\nSquare \(A
Rollout 10
pattern, n=22 is 22≡2 mod5, so losing. But let's
Rollout 8
157.\n\n**Final Answer**\n\boxed{157}\n\n<|im_start|>answer\nKathy
Rollout 0
}=128.\n\n**Final Answer**\n\boxed{128}\n\n<|im_start|>answer\nGiven a
Rollout 10
7 was losing. Check if n=14≡0 mod7. n=14: W
Rollout 10
0 (even), next is 2 (even), next 5 (odd), then 7
Rollout 10
18 is winning.\n\nTherefore, 22-4=18 (win). Subtract9→
Rollout 11
is not on unit circle either.\n\nAdditionally, (denote S = a + b + c, T
Rollout 5
the correct answer. \n\n**Final Answer**\n\boxed{181}\n\n<|im_start|>answer\nLet the
Rollout 10
, odd. Wait, starting from 0 (even), next is 2 (even), next
Rollout 13
7.12.\n\n**Final Answer**\n\boxed{187.12}\n\n<|im_start|>answer
Rollout 11
, 3)}.\n\n**Final Answer**\n\boxed{(-1, 3)}\n\n<|im_start|>answer\n
Rollout 2
{109}\n\n**Final Answer**\n\boxed{109}\n\n<|im_start|>answer\nTo find
Top 16 Negative Activations
Rollout 9
better way. Since each side is 10 units. So AI would be vertical, IM horizontal,
Rollout 9
a better way. Since each side is 10 units. So AI would be vertical, IM horizontal
Rollout 9
that's a better way. Since each side is 10 units. So AI would be vertical,
Rollout 12
t2 - ν2)^{-1/2} behaves like ( (t - ν)(t +
Rollout 0
20!\) and \(a < b\), so pairs (a,b) and (b
Rollout 8
example, how is the shuffling done? She shuffles the 10 cards (which are
Rollout 9
10,10). So these are straight lines. Let me find equations for GE and GM.\n\n
Rollout 12
2} [ dθ / ( dt/dθ ) ) ] ? Wait, we have determined that d
Rollout 12
+ (1 - ν2) sin2θ )\n\nHence,\n\nf(nu) = ��
Rollout 12
, the integral itself is convergent at x = 1/ν, so the derivative might still be
Rollout 11
any term can be expressed in terms of two variables. For example, let's write c = 1
Rollout 1
an infinite set where each pair is separated by d, perhaps S must lie on such a quadratic manifold.
Rollout 9
So the square is AIME with sides 10 units each. Then we have an isosce
Rollout 9
it's a square, all sides are 10 units. Let's label the square with points A
Rollout 9
would be horizontal, splitting EM into two parts of 5 units each. But also, since EM is
Rollout 9
's a square, all sides are 10 units. Let's label the square with points A,

DOWN_PROJ

Top 16 Positive Activations
Rollout 10
can have subtractions only up to s <=n. For each n, it's a finite number
Rollout 3
, temperature must be in Kelvin in the formulae.\n\nHence, given the confusion, perhaps the problem
Rollout 8
Wait, hold on. Wait, wait, no. If she is laying out 5 red cards,
Rollout 8
0. Wait, hold on. Wait, wait, no. If she is laying out 5 red
Rollout 0
to their exponents).. Wait, confusion arises here.\n\nWait, actually, we need to be precise.
Rollout 10
s) | s is a square, s<=n}.\n\nSo let's compute the Grundy numbers
Rollout 8
10. Wait, hold on. Wait, wait, no. If she is laying out
Rollout 0
of coprime ordered pairs, or is it?\n\nWait, but perhaps the counting is as follows:
Rollout 10
- s] | s is square, s <=n}\n\nSo mex is the smallest non-negative integer not
Rollout 8
out of 10. Wait, hold on. Wait, wait, no. If she is laying
Rollout 11
need to find all possible ordered pairs (p, q). Hmm, complex numbers with product 1.
Rollout 11
. Need to find possible ordered pairs (p, q).\n\nMaybe using the fact that if a complex number
Rollout 7
slightly ambiguous. Let me recheck the problem statement.\n\n“due to an error in the wording of a
Rollout 12
=nu2. Wait, no. Hold on. Wait:\n\n From earlier, the expression:\n\nE(k
Rollout 9
Or is the altitude segment inside the triangle? Wait. Wait, in an isosceles triangle,
Rollout 9
the altitude segment inside the triangle? Wait. Wait, in an isosceles triangle, the altitude
Top 16 Negative Activations
Rollout 3
8*R*T)/(π*M)), R is in J/(mol·K), which is same as (
Rollout 3
/(mol·K), which is same as (kg·m2)/(s2·mol·K
Rollout 10
In normal play convention, a position is losing if every move leads to a winning position for the opponent.
Rollout 10
positions (next player can force a win) or P-positions (previous player can force a win).
Rollout 10
there are infinitely many initial situations in which the second player can win no matter how his opponent plays.\n<|im_start|>
Rollout 8
126 is 2* 63=2*7*9. 31
Rollout 8
also not work. So we need to calculate the number of such "happy" sequences divided by the total
Rollout 8
/ 10P5\n\nAlternatively, compute numerator and denominator.\n\nFirst, compute denominator: 1
Rollout 3
, which is 8.314 J/(mol·K).\n\nAssuming standard temperature. Since
Rollout 10
the opponent. Winning position if there's at least one move leading to a losing position for the opponent.\n\n
Rollout 3
constant, which is 8.314 J/(mol·K).\n\nAssuming standard temperature.
Rollout 0
what about multiplicity. For example, 1 and 720 were co-prime (a
Rollout 10
is a losing position, as the player to move loses immediately. Then n=1 is a winning position
Rollout 7
original promotion. Therefore, N must be a common multiple of 3 and 12, that is
Rollout 10
. The player who is unable to make a move loses. Prove that there are infinitely many initial situations
Rollout 4
.000076996. sqrt(0.000076

Layer 26

GATE_PROJ

Top 16 Positive Activations
Rollout 3
Let's see, 168^2 = 28,224. 1
Rollout 3
:\n\n70*66.512=4,655.84\n\n3
Rollout 5
500) = sqrt(100*725) = 10*sqrt
Rollout 4
75 �� Compute 4975*1190=4975*(
Rollout 4
4975*1190=4975*(1200-
Rollout 4
,000 and 2000*15=30,000 and
Rollout 4
5025 * 1200 = 6,030,000
Rollout 8
's see: 240×30 =7200, 7440
Rollout 3
4\n\n3*66.512=199.536\n\nTotal=
Rollout 3
66.48*300=19,944 – 66
Rollout 5
= sqrt(100*725) = 10*sqrt(725
Rollout 5
8125) = sqrt(25 * 725) = 5 * sqrt
Rollout 4
015 = Compute 2000*3000=6,000
Rollout 3
*66.512:\n\n70*66.512=4,6
Rollout 3
66.512 =\n\n200*66.512=13,
Rollout 3
3= compute 66.512*290=19,288
Top 16 Negative Activations
Rollout 0
, numerator a and denominator b must satisfy that. However, a different way to phrase this is: a
Rollout 11
to both p and q being real.\n\nHence, maybe such a case is impossible, so perhaps we
Rollout 11
not real, which violates the problem conditions. Hence, such a choice of a=b, c=1
Rollout 7
. However, the problem states N<40. 36 is less than 40,
Rollout 9
would be outside the square.\n\nBut the problem states that the area common to triangle GEM and square A
Rollout 15
lamps will be off after some time".\n\nIn that case, the answer is n being given that are powers
Rollout 7
.\n\nSame result, which is impossible. Therefore, the conditions given in part(b) areimpossible.\n\n
Rollout 10
, there must be a mistake in my earlier assumption.\n\nTherefore, my pattern breaks at n=25
Rollout 13
and A_quarterly calculations once more because an error here could throw off the difference.\n\nA_annual
Rollout 0
be resolved.\n\nWait, perhaps that's the confusion. The original number is some fraction as part of between
Rollout 3
the rms speed. If they wanted the rms speed, but the question says "mean".\n\nAlternatively, maybe
Rollout 10
are losing. Therefore, there must be a mistake in my earlier assumption.\n\nTherefore, my pattern breaks
Rollout 3
" speed, it's probably the average speed formula. But maybe I should check. If I can't
Rollout 7
mistake in assumptions, perhaps the problem in part(b) is impossible? But no, the question is "
Rollout 0
0 factorial. Wait, that needs to be resolved.\n\nWait, perhaps that's the confusion. The original
Rollout 7
. The pass mark is still 65. Therefore, after the increase, to be promoted, you

UP_PROJ

Top 16 Positive Activations
Rollout 9
triangle GEM and square AIME is 80. Since the square is from (0,0
Rollout 5
(H - h) = 62.5H.\n\nSo we have two equations:\n\n1.
Rollout 5
x)/2 * h = 62.5H. The remaining trapezoid has bases
Rollout 9
\) and square \(AIME\) is 80 square units. We need to find the length of
Rollout 9
to both the triangle and the square is 80 square units, and I need to find the length
Rollout 4
(C1 + C2)) ] + dC2 [ (C1 + C2 - C
Rollout 8
total number of possible sequences is 30240.\n\nNow, we need to count how many
Rollout 4
\frac{\Delta C_{\text{eq}}}{C_{\text{eq}}} \right
Rollout 14
} \sqrt{A} + \frac{K}{A}.\n\]\n\nWe then minimize the right
Rollout 3
gas is \boxed{170} m/s.\n\nWait, after saying 168.
Rollout 9
triangle GEM and square AIME is 80 square units. The square area is 10
Rollout 9
inside the square (the common area is 80), and the square's right edge is EM,
Rollout 1
x⟩ + ⟨y, y⟩ - d2 / 2.\n\nBut this is a system of
Rollout 4
(C1 + C2)) ] + dC2 [ C1 / (C2 (C
Rollout 4
C_eq was 4.327 pF, C_eq is 1200 p
Rollout 4
1/(C1 + C2)] + dC2 [1/C2 - 1/(C
Top 16 Negative Activations
Rollout 3
n) in m/s\n<|im_start|>assistant\n<|im_start|>think\nOkay, I need to calculate the mean molecular
Rollout 13
value to the nearest cent.\n<|im_start|>assistant\n<|im_start|>think\nOkay, so Joanie took a $6
Rollout 4
this combination of capacitors?\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to find the percentage
Rollout 14
dx \right) $$ \n<|im_start|>assistant\n<|im_start|>think\nOkay, so I have to prove this inequality
Rollout 0
$ be the resulting product?\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I need to figure out how
Rollout 9
$\triangle GEM$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to solve this geometry
Rollout 13
to the nearest cent.\n<|im_start|>assistant\n<|im_start|>think\nOkay, so Joanie took a $6,
Rollout 2
1000$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I have this problem here:
Rollout 5
/100$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. I need to
Rollout 8
$m + n$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. We need to
Rollout 1
mathcal{H}$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to show that there
Rollout 11
.\n\n*David Altizio*\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I have this problem here where
Rollout 12
[P. Turan]\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to prove that the
Rollout 7
$47$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I'm trying to solve this
Rollout 4
combination of capacitors?\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to find the percentage error
Rollout 10
matter how his opponent plays.\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to prove that in

DOWN_PROJ

Top 16 Positive Activations
Rollout 4
990 * 2985= (2000 -10)*(30
Rollout 13
09\n\n1.03^4 = (1.0609)^2\n\nCompute
Rollout 3
3 * 66.512 = (200 +70 +3)*6
Rollout 3
is 300 - 2, so subtract 66.512 * 2
Rollout 13
\n\nThen 1.03^8 = (1.12550881
Rollout 2
mod1000 is109. Correct.\n\nFour terms:9*99*9
Rollout 4
)=4975*1200 -4975*10=5,
Rollout 4
0036055 squared: Approximately (0.0036)^2 =
Rollout 2
09*999=109*(1000-1)=109
Rollout 13
3600 = 9600, as above)\n\n6000 * 0
Rollout 13
, 1.12^4 would be (1.12^2)^2 = (
Rollout 10
3-25=8 (Grundy[8}=1), 33-16=
Rollout 3
sqrt(x)=168.5.\n\nSo sqrt(28,392.25
Rollout 3
4.\n\n168.5^2= (168 + 0.5)^2
Rollout 3
, which is 300 - 2, so subtract 66.512 *
Rollout 3
Let's see.\n\n168^2 = (170 - 2)^2 =
Top 16 Negative Activations
Rollout 3
on (Rn) in m/s\n<|im_start|>assistant\n<|im_start|>think\nOkay, I need to calculate
Rollout 15
i-1,t} + x_{i+1,t} mod 2\n\nThis is similar to
Rollout 11
= x, b = y, c = 1/(xy). Then, substitute into p and q
Rollout 12
certain way.\n\nAlternatively, I recall that in differentiation under the integral sign, if the integrand has a
Rollout 8
block and all the green cards are in a block. But since she's only laying out 5 cards
Rollout 0
^{}!$ be the resulting product?\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I need to
Rollout 14
�� f(y) dy * �� f(z) dz = ��∫∫ f(x)f
Rollout 4
energy stored in this combination of capacitors?\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to
Rollout 0
subset) and one unique b (multiplying the rest).\n\nSince each subset has a unique complement, the
Rollout 12
1$ . [P. Turan]\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to
Rollout 1
||2 - d2)/2.\n\nBut we don't know if all points in S have the same
Rollout 12
Wait, let's recall that when we use differentiation under the integral sign, as long as the integral conver
Rollout 11
=1 , which is prohibited since |c|=1 is disallowed. Thus, this also fails.\n\n
Rollout 3
which is related to temperature and molar mass. Wait, but the question specifically says "mean molecular speed
Rollout 11
symmetric approach where a and b are conjugates and c=1/(ab) didn't work. Trying
Rollout 11
. However, if I consider that abc = 1, we can make substitutions. Let's note that

Layer 27

GATE_PROJ

Top 16 Positive Activations
Rollout 5
.\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. I need to solve this problem about a
Rollout 8
.\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. We need to find the probability that when
Rollout 7
$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I'm trying to solve this math problem about participants
Rollout 0
product?\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I need to figure out how many rational numbers between
Rollout 13
cent.\n<|im_start|>assistant\n<|im_start|>think\nOkay, so Joanie took a $6,000
Rollout 2
$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I have this problem here: I need to find
Rollout 9
$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to solve this geometry problem where there's
Rollout 15
?\n<|im_start|>assistant\n<|im_start|>think\nAlright, let's try to figure out this problem. So, we
Rollout 14
$$ \n<|im_start|>assistant\n<|im_start|>think\nOkay, so I have to prove this inequality: \n\n$$\n
Rollout 11
izio*\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I have this problem here where a, b,
Rollout 3
in m/s\n<|im_start|>assistant\n<|im_start|>think\nOkay, I need to calculate the mean molecular speed of
Rollout 5
00$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. I need to solve this
Rollout 4
of capacitors?\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to find the percentage error in
Rollout 4
itors?\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to find the percentage error in the calculation
Rollout 10
plays.\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to prove that in this pebble game
Rollout 9
GEM$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to solve this geometry problem where
Top 16 Negative Activations
Rollout 0
count above). But 2^{3-1}= 4 correct. Hence, Therefore, the formula
Rollout 0
4, which is 2^{3-1}=4.\n\nSimilarly, if n is a square-free
Rollout 0
128. Alternatively, total number of subsets is 2^8=256, each
Rollout 9
would be (10 *10)/2 =50, so overlap area is 50
Rollout 2
^k mod8. But 2^3=8≡0 mod8, 2^k
Rollout 4
14.8058)/15 = 0.39/15 ��
Rollout 8
5*4*3 * 5*4= 60*20=120
Rollout 8
10*9*8*7*6≈7440 /30240
Rollout 4
, so total≈14,328 +477.8388≈1
Rollout 8
40= Let's check by dividing numerator and denominator by 240: 744
Rollout 0
20 is 4, which is 2^{3-1}=4.\n\nSimilarly, if n
Rollout 10
16.\n\nSubtract 1→14≡4 mod5, which is a winning position.
Rollout 0
problem. It's 2^{8-1}=128. Thus the answer is 1
Rollout 15
? Because n=1 is 2^1 -1 =1, and then maybe 3,
Rollout 0
. Since the total number is 2^8=256 (since each prime is either assigned
Rollout 8
0= Let's check by dividing numerator and denominator by 240: 7440

UP_PROJ

Top 16 Positive Activations
Rollout 2
8, so 10^k mod 8 is 2^k mod 8.\n\nThus
Rollout 2
(10^k -1) mod 8 is (2^k -1) mod
Rollout 2
Since 10 �� 2 mod 8, so 10^k mod 8
Rollout 2
^k -1) �� -1 mod8.\n\nTherefore, for k=1: (2
Rollout 2
8.\n\nCompute 2^k mod 8:\n\nNotice that 2^1 = 2
Rollout 9
+ (10 - y)^2]\n\nSince GE = GM, then:\n\nsqrt[h^2 +
Rollout 2
0 -1)= -1 ��7 mod8\n\nk=4: same as k=3
Rollout 2
1) = 2 -1 =1 mod8\n\nk=2: (4 -1)=
Rollout 7
originally scored 60 would now have 65, so they would be promoted. So some repe
Rollout 2
is (2^k -1) mod 8.\n\nTherefore, P mod 8 is the product
Rollout 1
2) } is not convex. Oh, right. The sphere is not convex. So perhaps closed balls
Rollout 2
that 2^1 = 2 mod 8\n\n2^2 = 4 mod 8
Rollout 2
8\n\n2^3 = 0 mod 8\n\n2^4 = 0 mod 8
Rollout 9
with GE = GM.\n\nCalculating GE distance:\n\nGE is from G (10 - h, y
Rollout 2
mod 8 is 2^k mod 8.\n\nThus, each term (10^k
Rollout 10
, let's think modulo4.\n\nSquares mod4 are 0 or1. Hence subtracting square
Top 16 Negative Activations
Rollout 2
, then combine the results using the Chinese Remainder Theorem.\n\n1. **Modulo 8 Calculation
Rollout 0
the product \(a \times b = 20!\).\n\n1. **Prime Factorization of
Rollout 10
2, 15, 17, 20, 22, 34
Rollout 3
not specify the temperature, we assume a standard temperature of \(298 \, \text{K
Rollout 3
:**\n - Compute the numerator: \(8 \times 8.314 \times
Rollout 7
so \( N \) must be a multiple of 3.\n\n2. **After score increase:**\n
Rollout 9
0)\), \(M(10, 10)\), and \(E(10,
Rollout 4
for capacitors in series is given by:\n\n\[\nC_{\text{eq}} = \frac
Rollout 7
= 2R \) and \( N = 3R \), so \( N \) must
Rollout 4
energy stored in two capacitors connected in series, we start by determining the equivalent capacitance \( C_{
Rollout 10
: 0, 2, 5, 7, 10, 12,
Rollout 2
+ 109\). Then:\n \[\n 125m + 1
Rollout 10
- Players take turns removing square numbers (1, 4, 9, 16, etc
Rollout 3
molecular speed of radon gas is \boxed{170} m/s.\n\nWait, after saying
Rollout 10
numbers (1, 4, 9, 16, etc.).\n - A player
Rollout 10
, 2, 5, 7, 10, 12, 15

DOWN_PROJ

Top 16 Positive Activations
Rollout 14
2.25≤8, so the inequality holds. But this seems that the ratio is 2
Rollout 0
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nGiven a rational
Rollout 3
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nCompute the mean
Rollout 1
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet $
Rollout 4
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nTwo capacitors
Rollout 6
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet's say
Rollout 8
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nKathy has
Rollout 15
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet $
Rollout 10
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nConsider the following
Rollout 11
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nSuppose
Rollout 9
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nSquare $A
Rollout 14
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nA function
Rollout 7
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nIn a mathematics
Rollout 12
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nProve that
Rollout 2
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nFind the remainder
Rollout 13
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nJoanie takes
Top 16 Negative Activations
Rollout 7
the new average of the promoted is 75 and non-promoted is 59. The total
Rollout 13
.00351936 * 6000 = 0.00
Rollout 2
. When k=1: 9 is odd. 10^k-1 in general is
Rollout 6
Problem 1:**\n\nTo construct the sparse set \( S \), we can combine the sparse sets \(
Rollout 6
classes. Let's start with the first one.\n\n**Problem 1:** Given k sparse sets S_
Rollout 11
instead of analysis. Suppose we note that given p and q are real, can we relate p and q
Rollout 7
10R\n\nDivide both sides by 5:\n\nP = 2R\n\nSo in
Rollout 7
, original score + 5 must be >=65? Wait, but if a student's original score
Rollout 0
terms, that means that a and b are coprime, hence exponents of primes in 2
Rollout 0
\((a, b)\) where \(a \times b = 20!\) and \(
Rollout 7
>=60 and <65. 98 not possible.\n\nN=24: x=
Rollout 10
12 to get 11 (winning), 12-4=8 (winning
Rollout 0
and \(a\) and \(b\) are coprime positive integers. The product \(a \times
Rollout 10
for the opponent. But for n=25, it's possible to take 25 pebb
Rollout 11
perhaps keeping variables as they are and using symmetries or equations from p and q.\n\nGiven that both
Rollout 10
(OEIS A017101?), but verification required.\n\nHowever, finding the specific positions

Layer 28

GATE_PROJ

Top 16 Positive Activations
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olar mass and R? No, the question is to compute the speed, so they must assume a temperature
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exist. As a contest problem, the solution may be found in realizing the periodicity failed, but that
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side is $I^3$, perhaps writing that as $I \cdot I \cdot I$, and
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20!.\n\nHence, the question is equivalent to: how many pairs (a, b)
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infinite number of n which are P-positions. To prove existence, perhaps without identifying them explicitly. That
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positions recur infinitely.\n\nAlternatively, considering that the set of square numbers has density zero in natural numbers, the
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subject.\n\nAlternatively, but, finally, the answer is asserting that such a y exists because he’s able
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that T is needed. But since the question is to compute, we proceed assuming 298 K
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confirm. But let's see, the question is asking for the "mean molecular speed v"—maybe the
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12,15,17,... which is 0,2,5,7,1
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∈S} is orthonormal.\n\nWhich is equivalent to requiring that { x - y } is orthogonal
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61.4 m/s.\n\nAs an assistant, I need to perhaps suggest both possibilities, but likely
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"the probability that Kathy will be happy", which is dependent on the arrangement of colors. So if there
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But the way the problem is phrased: "all the red cards laid out are adjacent and all
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specify the temperature. Hmm. But since they said to compute it, maybe we need to state standard assumptions
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equations in H for y.\n\nNow, the set of solutions to a single equation ||x - y||
Top 16 Negative Activations
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_{\text{quarterly}} = 6000 \times 1.604
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so OO maps to OO. That's a problem.\n\nWait, if we start with OO at minute
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:**\n - Total score: \( 71N = 75(P + x) +
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: \( 71N = 75(P + x) + 59(R - x
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: \( 71N = 79(P + x) + 47(R - x
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2N/3) + x ) +59*( (N/3) - x ) =
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75(P + x) + 59(R - x) \)\n - Substit
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A_{\text{quarterly}} = 6000 \times 1.60
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A_{\text{annual}} = 6000 \times 1.573
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8.314 \times 298 = 19,820.
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79(P + x) + 47(R - x) \)\n - Solving similarly
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{2000 \cdot 3000}{2000 + 30
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2^8 = 256\) coprime pairs \((a, b)\).\n\n3
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equiv 109 \mod 125\n \]\n\n3. **Combining Results
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47*(N/3 -x)=71N.\n\nLet's expand:\n\n79*(2
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times (-1) \equiv -891 \equiv -16 \equiv 109

UP_PROJ

Top 16 Positive Activations
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000, since 5025*1000=5,025
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e.g., k = 2i (modulus 2), then c =1/( (2
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c= e^{-i2θ}. But modulus of a and b is 1, which is prohibited
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). Let's see, 168^2 = 28,224.
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x} dx = 1/k^2$.\n\nThen the left-hand side is $(1/k)^
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2^4 would be (1.12^2)^2 = (1.254
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(10 - y)^2]\n\nSquaring both sides:\n\nh^2 + y^2
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628.23864 - 441.11616
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28,408.12-28,392.25=
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)= approximately 0.28125 <1, so inequality holds but again doesn't achieve
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1.4, as 161^2=25,921;16
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Then, their product is 1990*2985. Let's compute that.\n\n
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6.\n\nBreaking it down:\n\n6000*1 = 6000\n\n60
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do this step by step.\n1.12^1 = 1.12\n1.
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I≈1.31, then I^3≈(1.31)^3≈2
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25:\n\nFirst, 5025 * 1200 = 6,0
Top 16 Negative Activations
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[ t / sqrt( (t^2 - ν^2)(1 - t^2) )
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= [ t / sqrt( (t2 - ν2)(1 - t^2) ) ]
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_{t+1}=x2_t, without mod necessarily.\n\nSimilarly, for internal lamps with two neighbors
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sqrt( (1 - t2)/(1 - ν2) )\n\nTherefore, sinθ cosθ =
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],\n\n [0,0,0],\n\n [0,1,0]]\n\nThen, M^
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,10), (0,10 -50/h), (0,50/h),
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over the integers (without modulus), but then immediately reducing each entry modulo 2 (since the next state
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[ E(k) ] / ( (1 - nu2) nu ) + [ nu2 K(k
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) \boxed{12}, \boxed{24}, \boxed{36}\n\n(b)
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0), (0,10 -50/h), G's projection path, etc. Wait,
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, dt / [ sqrt( (t2 - ν2)(1 -t2) ) ] =
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0/h) and (0,10 -50/h), which is between (0, approximately
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) = (1 - t2)/(1 - ν2)\n\nTherefore, cosθ = sqrt( (
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/h) and K (0,10 -50/h).\n\nBut 10 -50
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sqrt(A)=sqrt(2K/sqrt( J ))}= (2K/ J^{1/
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modeled: since actual Lamp states do not involve any mod2 operations, their next state depends strictly on neighbors

DOWN_PROJ

Top 16 Positive Activations
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109 in both cases. So The answer is109. Huh.\n\nBut wait,
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] + [ r2 + 2/r cosθ ]= 2r cosθ +1/r
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.246.\n\nThus, the answer would be 31 +126=15
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, a * a z *1/(a2 z )=1, which holds.\n\nRegardless, the
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ν)^{-1/2}, but when integrated, since we have (t - ν)^{-1
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-1}=128. Thus the answer is 128. Hence, the answer is
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�� must satisfy cosθ=-2/5. However, �� = nπ/3.\n\n
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conjectured to be false. Thus, the answer is P = P_bad-angel, and NP ≠
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20! is not just the product of primes, but the product of primes raised to various powers.
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) by Cauchy-Schwarz, but as before, ��1^2 dx is infinite
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is impossible in polynomial time.\n\nSo perhaps the answer is not straightforward, hence why reference to the P/p
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128 such numbers. Therefore, The answer is 128, corresponding to \boxed{
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For k=3:\n\nSame as k=2 but reversed. Because k=3 reds and
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R = 66P + 66R\n\nSubtracting 66P and
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� (f(x))^2 x^a dx )^{1/2} (�� x^{-a
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1 has modulus 1 and argument 0, so modulus r_a r_b r_c =
Top 16 Negative Activations
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2) d �� 1.414d. The mutual distance between centers is d.
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(10 -5)/(10 - (10 - h)) = 5/h.\n\nTherefore
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sequences must be either R^r G^{5−r} or G^{5−r} R
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winning position for the opponent. Winning position if there's at least one move leading to a losing position for
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.00351936 * 6000 = 0.00
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wording is slightly ambiguous. Let me recheck the problem statement.\n\n“due to an error in the wording
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). So sqrt(8)=2.828. 475 /2.82
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. Wait that's grouped. 0,2; 5,7;10,12
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at least one move leading to a losing position for the opponent.\n\nSo starting from n=0: losing
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2, every position is W. But that seems wrong, because as n increases, there's more squares
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they use 273 K, others 298 K. For example, if it's
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, we need to consider all possible partitions where there's at least one red and one green. Also,
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seems not immediately helpful. \n\nWait, have you consider applying the Cauchy-Schwarz inequality
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0.\n\nWait, but hold on. Let me check for k=3.\n\nWait, k=3
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(x) being integrable. Let me think about if F(x)=∫_{x}^\infty
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22, every position is W. But that seems wrong, because as n increases, there's more

Layer 29

GATE_PROJ

Top 16 Positive Activations
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approach. Wait, maybe I should try direct multiplication:\n\n1.2544\n*1.
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BB... Wait, GGRR would actually be similar: P(5,2)*P(5,
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dx\n\nWait, wait subtract the lower from upper:\n\n= ��[0 to10] [
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on and off as integers without mod2 consideration, starting from O O O:\n\nt0: O O
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6000 = ?\n\nWait, maybe better to process 6000 * 1
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. Maybe this will increase precision?\n\nWait, or rather:\n\nWait, before the coefficient for ΔC1
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by squaring or logarithms. Oh, maybe better to compute step by step.\n\nWe can also use
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lie on such a sphere. Wait, no. What we need is more along the lines of for each
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EA. Wait, maybe that's a better way. Since each side is 10 units. So
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1936\n\nWait, or perhaps it's easier to multiply directly. Still, this might take
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002 *10, etc. Let's recalculate the derivatives values more precisely:\n\nStarting with
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But wait, this might be another approach. Let's see:\n\nIf I take the natural log of C
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* λ^{-?} Wait sorry scaling variables properly:\n\nOriginal variables:\n\nLet set y= λ x.
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using Hölder’s for three variables. Wait, another approach.\n\nAlternatively, maybe use Hölder’s inequality
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, not the best approach. Wait, maybe I should try direct multiplication:\n\n1.2544
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-r))! ; wait, maybe it's better to think:\n\nWait, arranging r reds and
Top 16 Negative Activations
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cal{H}$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to show that there exists
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P. Turan]\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to prove that the function
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) in m/s\n<|im_start|>assistant\n<|im_start|>think\nOkay, I need to calculate the mean molecular speed
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\right) $$ \n<|im_start|>assistant\n<|im_start|>think\nOkay, so I have to prove this inequality:
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how his opponent plays.\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to prove that in this
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000$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I have this problem here: I
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combination of capacitors?\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to find the percentage error
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to the nearest cent.\n<|im_start|>assistant\n<|im_start|>think\nOkay, so Joanie took a $6,
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be the resulting product?\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I need to figure out how many
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$47$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I'm trying to solve this math
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[P. Turan]\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to prove that the
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. [/*]\n<|im_start|>assistant\n<|im_start|>think\nOkay, let me try to work through these problems
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off after some time?\n<|im_start|>assistant\n<|im_start|>think\nAlright, let's try to figure out this problem
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*David Altizio*\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I have this problem here where a
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m + n$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. We need to find
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100$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. I need to solve

UP_PROJ

Top 16 Positive Activations
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b\). So for each such coprime pair where \(a < b\), we have a
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2?\n\nAlternatively, substitution φ where t = cosθ? Let me think. For if t is from
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and \(b\) are coprime, each prime factor in 20! must go entirely to
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!, where \(a\) is the product of the primes in the subset and \(b\) is the product
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0, but they are 1 and the complex roots. But those have modulus 1. Not allowed
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subset of the prime factors of 20!, where \(a\) is the product of the primes in
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ime.\n\nHence equivalence between choices over subsets of primes.\n\nHence, since ordered coprimes,
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1. So discard.\n\nPerhaps substitute three numbers as solutions to equations like x^3 -qx^2
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2^8=256 (since each prime is either assigned to a or to b,
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integral representation of the elliptic integral of the first kind. Let me check.\n\nIn the standard form of
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n, you can't encode solutions for all possible formulas of size n because SAT is NP-complete. Thus
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{ y + (d / sqrt(2})) e_x } where {e_x} is orthorn
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squares (k2 and (k+1)2), the possible moves that can be made are the
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problem reduces to finding the number of coprime pairs \((a, b)\) where \(a
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is 71, and the average of those who failed (repeaters) is 56.
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applies; each such pair is obtained by assignment of primes to a or b, since a and b need
Top 16 Negative Activations
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\nu < 1 \).\n\n\[\n\boxed{f(\nu) \text{ is mon
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5 is 181.\n\n\[\n\boxed{181}\n\]<|im_end|>
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= 157\).\n\n\[\n\boxed{157}\n\]<|im_end|>
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7.12\n\]\n\n\[\n\boxed{187.12}\n\]
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answer is 25. Therefore box form \boxed{25}\n\nWait, that seems correct.\n\n
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Thus, the inequality is proven:\n\n\[\n\boxed{ \left(\int_0^\infty
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6= 157.\n\n**Final Answer**\n\boxed{157}\n\n<|im_start|>answer\n
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answer is 128, corresponding to \boxed{128}? Or wait, wait
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as per problem statement.\n\n**[Final Answer]**\n\boxed{\text{Such a point } y
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157.\n\n**Final Answer**\n\boxed{157}\n\n<|im_start|>answer\nKathy
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26= 157.\n\n**Final Answer**\n\boxed{157}\n\n<|im_start|>answer
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\boxed{109}\n\n**Final Answer**\n\boxed{109}\n\n<|im_start|>answer\n
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are powers of two. The final answer is:\n\n\[\n\boxed{2^k}\n\]\n\n
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the correct answer. \n\n**Final Answer**\n\boxed{181}\n\n<|im_start|>answer\nLet the
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{169} m/s.\n\n<|im_start|>answer\nTo compute the mean molecular speed \( v \)
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the proof is complete.\n\n**Final Answer**\n\n\boxed{f(\nu) \text{ is mon

DOWN_PROJ

Top 16 Positive Activations
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the 1/x integral between \ the tails.\n\nAlternatively, use Hölder inequality in the form:\n\n(
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Alternatively, if the matrix is a Jordan block.\n\nAlternatively, we search for nilpotent adjacency matrices of
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But not exactly.\n\nAlternatively, taking reciprocal vectors.\n\nAlternatively, this can't take too long, given that
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roots of x^3 -1=0, but they are 1 and the complex roots. But
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1 as some function involving x and f(x).\n\nAlternatively, maybe use Hölder's inequality such that �
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. Not allowed. Or if variables are scaled.\n\nAlternatively, taking a complex number ��, non-real
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qx + r =0... But not exactly.\n\nAlternatively, taking reciprocal vectors.\n\nAlternatively, this can't
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||1||2 ||1||infty, but i think triple Hölder not in that manner.\n\n
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( ��x f(x)^2 dx ). But yet again the first integral diverges. However,
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b} etc. But modulo abc=1.\n\nAlternatively, take a = \overline{1/b
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all accepting paths, which might not be helpful.\n\nAlternatively, recall that P_angel is equivalent to P
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: pebbles equal to some composite structure, but I don't see how.\n\nAlternatively, an example
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px^2 + qx + r =0... But not exactly.\n\nAlternatively, taking reciprocal vectors.\n\nAlternatively
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: in base5, the digits are limited.\n\nAlternatively, look for mathematical theorems on such games
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(x)) dx, but this complicates the expression. Not sure.\n\nAnother thought: the given integrals
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)/2 + t^2 /2. But not sure.\n\nWait, perhaps optimize the ratio.\n\nWait
Top 16 Negative Activations
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Wait, but I might have to verify which one is the actual mean speed. Let me think.\n\nMean
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zero. So, the orthogonality condition comes automatically from the equidistant condition. So that
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50, as before. So how can we have overlapping area of 80? Since the square
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8125) which is exactly what we have. Thus, that is consistent.\n\nTherefore, x
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18125) which is exactly what we have. Thus, that is consistent.\n\nTherefore,
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(18125) which is exactly what we have. Thus, that is consistent.\n\nTherefore
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single block each.\n\nSo that makes sense. So how else can we model this?\n\nPerhaps treating the sequences
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x)\n\nSo the equation becomes:\n\n(x + 175)(x - 50)/(7
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the previous result.\n\nTherefore, combining this with the 2*0.004 (from the
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, so triangle is within. But then how come the overlapping area is 80 then? Because that
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, for color arrangements where all reds are together and all greens are together, regardless of the order (
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actually, we need to be precise. When we say that a and b are coprime with a
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F. C_eq=(2000*3000)/(2000+3
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Therefore, A=42 +56=98.\n\nBut that's impossible as A must be
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product over the sum, so C_eq = C1C2/(C1 + C2). The
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is dead end.\n\nThis is perplexing. So using symmetric approach where a and b are conjugates and

Layer 30

GATE_PROJ

Top 16 Positive Activations
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brackets to be real, require that t +1/t is real, which for complex t would imply that
Rollout 11
. Because:\n\nLet’s prove that t +1/t is real �� either t is real or |
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t1 =a, then [a +1/a ]=t1 +1/t1. Similarly
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sure. Let me set x =a +1/a, y =b +1/b. Then,
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and similarly for others. So a + 1/a = a + bc; similarly for the others.
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base \(EM\), and the area common to triangle \(GEM\) and square \(AIME\)
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), \(\frac{1}{b} = ac\), and \(\frac{1}{c
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\(\frac{1}{b} = ac\), and \(\frac{1}{c}
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x and y like with x = a +1/a and y= b +1/b, and maybe
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c/b^2 + a^2/(b c) + b^2/(a c) +
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for k=2, the color sequences are RRGGG and GGGGRR. Wait, RR
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p is given by p= r2 +1/r2 + 2 cosθ ( r +1
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cosθ + 1/r2\n\nab + bc + ca = (r e^{iθ})(
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equation 2 cosθ = r3 +1/r3.\n\nNote that if R = r3 +
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1/t1 ) + (t2 +1/t2 ) + (t1 t2 +1
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GGGGR\n\nk = 2: RRGGG, GGGGRR\n\nWait, wait
Top 16 Negative Activations
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of radon gas. Hmm, first, I remember there's a formula for the root mean square speed
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out how to approach this. \n\nFirst, I remember that a function is monotonically decreasing if its
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22 g/mol, right?\n\nSo, first, M = 222 g/mol =
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here.\n\nHence, you just reference this result. However, since the problem asks to show it,
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of 222 g/mol. Therefore, the molar mass M is 222 g
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figures. The question might not require rounding, but to check if my initial error propagations are correct.\n\n
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{-1/2} ] dt\n\nThen, to compute f’(ν), perhaps using Leib
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222 g/mol, right?\n\nSo, first, M = 222 g/mol =
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/100. \n\nFirst, let me recall some trapezoid properties. In a tr
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think\nOkay, let's see. I need to solve this problem about a trapezoid with
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6*0.222) )\n\nFirst compute numerator:\n\n8*8.314 =
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that's probably the answer here.\n\nHence, you just reference this result. However, since the problem
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root mean square speed. But maybe in this context, they just want the root mean square speed? The
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isotopes like Rn-220 and Rn-219. But the most common
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different from the root mean square speed. But maybe in this context, they just want the root mean square
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) becomes 59.\n\nPart (a) is asking for all possible values of N given these conditions

UP_PROJ

Top 16 Positive Activations
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for k>=3, hence (10^k -1)≡-1 mod125
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, since energy is (1/2)*C_eq*V2, maximum energy would be when C
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iz formula is -F(ν, ν) * 1, which is -∞, the actual
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the first term in Leibniz formula is -F(ν, ν) * 1,
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1*first row M^3 + 0*second + 1*third row M^3
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) is non-negative: at x=0, f(0)=c; at x=c/d,
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25 for k>=3, hence (10^k -1)≡-1 mod1
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ab = a b^2; and (1/(ab))/a is 1/(a^2
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, since a is a number composed by assigning primes to a (possibly a = product of primes raised to
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M * M^3:\n\nFirst row: 0*M^3 rows + 1*[0,
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e^{iθ}\n- c = 1/(r2 e^{i2θ})\n- ab
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xy) = x y^2\n\n(1/(xy))/x = 1/(x^2
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rule is that each lamp at t+1 is on iff exactly one neighbor is on at t. So
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because we can ignore the angel string (set α_n to some fixed string, which can be computed trivial
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^3:\n\nFirst row: 0*M^3 rows + 1*[0,0,0
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the lamp i, next state is (left neighbor state) XOR (right neighbor state). If lamp i
Top 16 Negative Activations
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, EM is the top side of the square. Wait, hold on. \n\nWait, in the square
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/b + a/c is the reciprocal of q? Wait, if we take reciprocal of each term in q
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5, that's original score >=60. Wait, but originally, the pass mark was 6
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1936.\n\nWait, that seems familiar but let me verify using another approach.\n\nSince 1
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+ (c + 1/c). Wait, wait a second. Since 1/a = bc,
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, so wait, that can't be right. Wait, actually multiplying 1.12 by
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5,7,10,12,...\n\nBut that's equivalent to numbers congruent to 0
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c/b + a/c is the reciprocal of q? Wait, if we take reciprocal of each term in
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are AI, IM, ME, and EA. Wait, maybe that's a better way. Since each
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8%. Wait, let me check the exact value.\n\nFirst, compute (ΔC_eq / C_eq
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is 2, which is even. Wait a second. Then primes 11, 13
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/b) + (c + 1/c). Wait, wait a second. Since 1/a =
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to 999. Wait, no. Wait, the first term is k=1 (1
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* 8 * 7 * 6. Alternatively, that's 30240.
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which rounds to 0.88%. Wait, let me check the exact value.\n\nFirst, compute
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31/126.\n\nWait, wait: 7440 ��240

DOWN_PROJ

Top 16 Positive Activations
Rollout 14
) sqrt(A ) +K/A.\n\nSet derivative with respect to A to zero.\n\ndh/dA=
Rollout 14
fixed $J$ and $K$. If I can show that the maximum possible value of $I^
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with exponents that are conjugate. Let me recall Hölder's inequality: for $p$ and
Rollout 9
h=10 would be (base * height)/2 = (10 * 10)/
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2x2) ). Let me set a substitution so that the upper limit becomes fixed. For example,
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of power functions. But not sure. Let me recall the Beta function:\n\nB(p, q) =
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ence, [0.003]^2 = 9e-6\n\nSecond term: [
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1/q +1/r =1. If we can split f, such that when multiplied with some other
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that creates a certain area ratio. Let me also recall that the area of a trapezoid is
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2000*3000 = 6,000,000
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to use the parallelogram law. Let me recall that in a Hilbert space, given two vectors
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)* sqrt(A)= sqrt(J)* (2K / sqrt(J))^{1/2 }= sqrt
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which is 80.\n\nTherefore, need to find h such that the area of overlap is 8
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is a square), leaving 0. Second player loses. So n=1 is a winning position.\n\n
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tail integral of f starting at x. If we can relate I to F and K.\n\nNo, that
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's see: 240×30 =7200, 7440
Top 16 Negative Activations
Rollout 8
adjacent). But the way the problem is phrased: "all the red cards laid out are adjacent
Rollout 5
= 0\n\nSimplify:\n\ny^2 - 150y - 125
Rollout 0
, which is odd. Actually, wait nvm, the exponents for 2,3,5
Rollout 9
from both sides:\n\n0 = 100 - 20y\n\nSo 20y
Rollout 14
1/(2k)$,\n\n$K = \int_0^\infty x e^{-k x
Rollout 15
^4 is zero matrix? Wait, no.\n\nWait, first row of M^4 is:\n\nRow
Rollout 14
= 1/k$,\n\n$J = \int_0^\infty e^{-2 k x
Rollout 10
mod5 are losing. Hmm. Wait. Wait, that is a problem. So in the initial thinking
Rollout 11
Thus, y=0 or x2 + y2=1. So y=0 (t is
Rollout 14
J = �� f2 dx and K=∫xf dx. A linear combination of J and K
Rollout 11
, and none have absolute value 1. They define p as (a + b + c) +
Rollout 0
9 <80: yes.\n\n4. Assign 5 to a, others to b: a
Rollout 0
16 <45: yes)\n\n3. Assign 3 to a =>a=9, b
Rollout 3
temperature, this seems ambiguous. However, in many standard problems, they might take 273 K
Rollout 10
5 is a losing position? Wait, no. Wait, according to the previous calculation:\n\nn=2
Rollout 7
2R\n\nSo in the original scenario, the number of promoted participants is twice the number of repeaters

Layer 31

GATE_PROJ

Top 16 Positive Activations
Rollout 13
than 1.5625.\n\nBut the exact calculation gives around 1.573
Rollout 2
09. But wait, but I feel like in this case, the product of all these numbers,
Rollout 8
's 30240. But let me verify that. Yes, 10 choices for
Rollout 8
: 1200\n\nBut wait, the total for each k is 2 * [P
Rollout 3
Wait, I need to confirm. But let's see, the question is asking for the "mean molecular
Rollout 10
W\n\nWait, when does the next L occur? Wait at n=26: subtract16
Rollout 3
a standard temperature to assume. Let me check online, but since I can't do that, I need
Rollout 9
0 and x=10). Wait, but in trapezoid area formula:\n\nAverage of the
Rollout 0
.\n\nBut wait, can it be? Let me check with a smaller factorial.\n\nSuppose instead of
Rollout 3
which one is the actual mean speed. Let me think.\n\nMean speed, or average speed, is indeed
Rollout 8
and green cards to the green positions. Let me see.\n\nFor a specific color sequence, say RRGG
Rollout 0
20! has exponents:\n\nLet me do prime factorization of 20!.\n\nTo compute
Rollout 3
radon gas at a standard temperature. Let me think. If I remember, for example, for gases
Rollout 8
of permutations (10P5).\n\nLet me try this approach.\n\nFirst, how many color sequences are
Rollout 2
09 mod1000. But let me check this process again.\n\nFirst modulus8:\n\nProduct
Rollout 0
divide the total by 2.\n\nWait, but the problem says "how many rational numbers between 0
Top 16 Negative Activations
Rollout 6
{P}_{angel}$ if there exists a polynomial $p : \mathbb{N}
Rollout 10
we analyze the game where players alternately remove a square number of pebbles. The key is to
Rollout 10
bbles where $x$ is the square of any positive integer. The player who is unable
Rollout 10
the game where players alternately remove a square number of pebbles. The key is to identify losing
Rollout 10
the Game**:\n - Players take turns removing square numbers (1, 4, 9,
Rollout 6
}$ with a constraint that there should exists a polynomial time algorithm that can **compute** the angel string
Rollout 10
analyze the game where players alternately remove a square number of pebbles. The key is to identify
Rollout 10
understand the game first.\n\nPlayers take turns removing a square number of pebbles (1, 4
Rollout 6
\{\alpha_n\} \) and a polynomial-time machine \( M \). By the Meyer's
Rollout 6
said to be **sparse** if there exists a polynomial $p : \mathbb{N}
Rollout 5
the midpoints of the legs, has a length equal to the average of the two bases, \( b
Rollout 10
to find sets where for any move (subtract a square) from a number in the set, you end
Rollout 10
Game**:\n - Players take turns removing square numbers (1, 4, 9,
Rollout 6
/poly is equivalent to languages decidable by a polynomial-time Turing machine with a polynomial-length advice string,
Rollout 10
(P-positions) where every move leads to a winning position (N-position) for the opponent.\n\n
Rollout 6
N}$ . In other words, there is a poly-time algorithm $A$ such that

UP_PROJ

Top 16 Positive Activations
Rollout 14
the maximum possible ratio is approximately 2.25/8≈0.28. Wait,
Rollout 14
But this seems that the ratio is 2.25/8=0.2812
Rollout 14
but the maximum possible ratio is approximately 2.25/8≈0.28. Wait
Rollout 14
this seems that the ratio is 2.25/8=0.28125
Rollout 14
)/(4/(k^3 ))=1/4. Which matches the ratio here. Hmmm.\n\n
Rollout 14
5/8=0.28125. Which indicates that equality is achieved in other cases
Rollout 14
2)= approximately 0.28125 <1, so inequality holds but again doesn't
Rollout 14
So ratio LHS/RHS= (1/8)/(4/9)=9/(32)=
Rollout 3
Rn, which is approx. 1/8 the speed (mass is ~8 times higher,
Rollout 7
(R - x).\n\nTherefore, A*x + B*(R - x) = 56R
Rollout 9
80, so maybe the triangle is partially outside and partially inside. If point G is outside the square
Rollout 9
(1/5 x +8 +1/5 x -2 ) ] dx\n\n= ��
Rollout 15
1 is good, n=2 and 3 are bad.\n\nNext, n=4. Let's
Rollout 14
64= 9/4=2.25.\n\nThen 8 J K=8*
Rollout 7
P=2N => P = (2/3)N.\n\nSo N must be divisible by
Rollout 12
So E(k) > (1 -k2) K(k). Let me check for some value.
Top 16 Negative Activations
Rollout 9
25.\n\nPerfect.\n\n**Final Answer**\n\boxed{25}\n\n<|im_start|>answer\nSquare \(A
Rollout 15
superdiagonal and first subdiagonal.\n\nWait, example for n=4:\n\nM would be:\n\n
Rollout 11
, 3)}.\n\n**Final Answer**\n\boxed{(-1, 3)}\n\n<|im_start|>answer\n
Rollout 5
the correct answer. \n\n**Final Answer**\n\boxed{181}\n\n<|im_start|>answer\nLet the
Rollout 3
69 m/s.\n\n**Final Answer**\n\boxed{170} m/s\n\nWait no
Rollout 0
}=128.\n\n**Final Answer**\n\boxed{128}\n\n<|im_start|>answer\nGiven a
Rollout 2
109.\n\nTherefore, boxed answer is \boxed{109}\n\n**Final Answer**\n\
Rollout 10
, the second player can win in these cases.\n\n**Final Answer**\n\boxed{There are infinitely many
Rollout 2
{109}\n\n**Final Answer**\n\boxed{109}\n\n<|im_start|>answer\nTo find
Rollout 8
permutations, so 10P5. Yeah.\n\nBut perhaps another perspective: if she's selecting
Rollout 8
wait, maybe it's better to think:\n\nWait, arranging r reds and 5−r greens
Rollout 8
s and 4 to 1 greens).\n\nWait, so for each possible k (number of reds
Rollout 12
k2 cos2θ\n\nWait, hold on:\n\nWait, let's write nu2 + (1
Rollout 5
181 is the correct answer. \n\n**Final Answer**\n\boxed{181}\n\n
Rollout 5
5 is 181.\n\n\[\n\boxed{181}\n\]<|im_end|>
Rollout 2
divided by 1000 is \(\boxed{109}\).<|im_end|>

DOWN_PROJ

Top 16 Positive Activations
Rollout 7
32x=8N/3 => x=(8N)/(3*32)=N/
Rollout 5
(x + 175)H [1 - 125/(75 + x)]
Rollout 10
considering that between consecutive squares (k2 and (k+1)2), the possible moves that can
Rollout 11
+ y2 )=0 → y [1 -1/(x2 + y2 ) ]=
Rollout 10
W if there exists a square number s <=n such that n - s is L.\n\nn is L
Rollout 14
= sqrt(J) * sqrt(2K)/ J^{1/4}= J^{1/2
Rollout 10
large k, the difference between k2 and (k+1)2 is 2k+1
Rollout 14
/3}/4.\n\nNote that 12=22 *3, so 12^{
Rollout 5
5H\n\nBut from the first equation: h = (125H)/(75 + x
Rollout 10
between consecutive squares (k2 and (k+1)2), the possible moves that can be made
Rollout 11
r3 +1/r3, then cosθ = R /2.\n\nBut also, since -1
Rollout 9
20=500/h\n\nThen h=500/20=25.\n\n
Rollout 10
to a losing position).\n\nn=4: from 4, you can move to 4-1
Rollout 14
) * A= 2K.\n\nThus A=2K / sqrt(J).\n\nPlug this into upper
Rollout 11
example, if abc = 1, then c = 1/(ab). Let me try that.
Rollout 10
, considering that between consecutive squares (k2 and (k+1)2), the possible moves that
Top 16 Negative Activations
Rollout 15
not cycling. Therefore, my mistake was introducing that previously. So maybe n=3 isn't cyclic.\n\n
Rollout 3
in kinetic theory, they are different. Let me make sure.\n\nAccording to Maxwell-Boltzmann distribution
Rollout 4
approximately 0.36%, consistent with the previous result.\n\nTherefore, combining this with the 2
Rollout 15
all on, ends after two steps. Then why did I earlier thought it cycles?\n\nBecause I made an
Rollout 14
) dx$, perhaps express it as integral over x from 0 to ��, and relate it to
Rollout 5
= 181.25, so the greatest integer not exceeding this is 181
Rollout 10
, that is a problem. So in the initial thinking, n=5 and n=10 were
Rollout 2
)≡109 mod125. Correct.\n\nThen solve for x≡5 mod8 and
Rollout 15
on, ends after two steps. Then why did I earlier thought it cycles?\n\nBecause I made an error
Rollout 5
now, but maybe I had a miscalculation in the algebra step. Let me verify quickly.
Rollout 2
the same final result, so now, why did my modulus8 and modulus125 approach via CRT
Rollout 1
/ 2 = 0.\n\nCompare to the above line:\n\nFrom the sum of norms, we have
Rollout 2
to modulo8 analysis? Let's be precise.\n\nbreak down the terms in the product when mod12
Rollout 15
t=2, not cycling. Therefore, my mistake was introducing that previously. So maybe n=3
Rollout 15
the system dies at t=2. But in my initial integer-based simulation (where I wasn't doing
Rollout 15
cycling. Therefore, my mistake was introducing that previously. So maybe n=3 isn't cyclic.\n\nWait

Layer 32

GATE_PROJ

Top 16 Positive Activations
Rollout 6
\) unless \( \textbf{NP} = \textbf{P} \).\n\n---\n\n**
Rollout 2
91 mod125: 125*7=875. 89
Rollout 12
})\n\]\n\nTo determine the monotonicity, we compute the derivative \( f'(\nu) \
Rollout 4
{C_1 \cdot C_2}{C_1 + C_2}\n\]\n\nGiven
Rollout 2
91. 891 mod125: 125*7=87
Rollout 2
9's}}$ is divided by $1000$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright
Rollout 12
2 - \nu^2)(1 - t^2)}}\n\]\n\nNext, we recognize this
Rollout 4
eq}} \) and its uncertainty. The equivalent capacitance for capacitors in series is given by:\n\n
Rollout 2
91: 891mod125. 125*7=87
Rollout 2
=999≡-1 mod125.\n\nIn fact, as noticed before, 1
Rollout 4
10 pF and C2 = 3000 ± 15 pF, and
Rollout 12
[\nf'(\nu) = -\frac{E(\sqrt{1 - \nu^2
Rollout 6
\), unless \( \textbf{NP} = \textbf{P} \), this equality
Rollout 7
0. 36 is less than 40, so all three are acceptable.\n\nWait, so
Rollout 12
1)(1 - \nu^2 x^2)}} \) is monotonically decreasing for
Rollout 8
= 10! / (10-5)! = 10 * 9 *
Top 16 Negative Activations
Rollout 0
1.\n\nSo primes 2: 18, 3: 8, 5:4
Rollout 4
significant figures, so 0.88%, or maybe the answer is exact at 0.8
Rollout 10
If n ��0 mod4: subtract0 or1.\n\n subtract s≡0 or1 mod4
Rollout 11
Hmm, given that p and q are symmetric in a, b, c in some way. p is
Rollout 3
is 168.5 m/s, and rounded to three sig figs is 16
Rollout 0
corresponding to \boxed{128}? Or wait, wait prime factors of 20!
Rollout 0
So exponents are: 18 (2), 8 (3), 4 (5),
Rollout 4
Then total relative variance in C_eq is 9e-6 + 4e-6 =
Rollout 14
compute this. Expand (c - dx)^2= c^2 -2c d x + d
Rollout 3
atomic number is 86, so atomic weight is approximately 222 (since the most stable
Rollout 9
also, since EM is vertical, G is somewhere to the left or right of EM. Since the triangle
Rollout 1
(d, squared + ||y|| squared ) / 2 = (d2 + d2 /
Rollout 2
97} mod125\n\nBecause from k=3 to k=999, that
Rollout 0
\(a \neq b\). Hence, total coprime pairs are \(2^{8}\
Rollout 15
1 (off) and L3 (on) at t=1, so exact one neighbor on.
Rollout 3
168.5≈ 168 or 169. In technical terms,

UP_PROJ

Top 16 Positive Activations
Rollout 3
*R*T has units (kg·m2)/(s2·mol). Then, dividing by M in
Rollout 1
non-empty?\n\nThere's a theorem in the theory of Hilbert spaces called the Kakutani's theorem
Rollout 6
= O(log n), how many strings of length m are in S_L? Since m is log n
Rollout 1
if we can identify y as the center of such a system. The problem is the existence of y.\n\n
Rollout 14
(x) is exponential, I^3/(J K)= (1/k^3)/ ( 8
Rollout 6
p(n), that corresponds to one string of length m per n equal to p^{-1}(m).
Rollout 6
_i has at most p(n) strings of length n. Since there are k sparse sets, maybe we
Rollout 13
_annual and A_quarterly calculations once more because an error here could throw off the difference.\n\nA
Rollout 14
, in fact, not tight. Wait, but maybe I had an error in Calculus of Variations
Rollout 1
. If S is a set of points each lying on the sphere of radius d/sqrt(2)
Rollout 1
\{ \frac{\sqrt{2}}{d}(x - y) : x \in S
Rollout 6
overlapping). Hence, the number of strings of length m is O(1), which is polynomial (constant
Rollout 6
Answer:** \n\( \textbf{P} = \textbf{P}_{\text{bad
Rollout 6
a polynomial value, the number of strings of length m is at most 1, which is sparse.
Rollout 8
30240 ��240= 126. So 31
Rollout 1
each pair is separated by d, perhaps S must lie on such a quadratic manifold. But unless I can
Top 16 Negative Activations
Rollout 11
:\n\nGiven abc=1, then taking reciprocals, (1/a)(1/b)(1/c
Rollout 11
's denote x = a, y = b, z = c. Then, given xyz = 1
Rollout 11
only.\n\nAlternatively, maybe take conjugate of variables. Let’s set x = \overline{a
Rollout 11
ate of variables. Let’s set x = \overline{a}, y = \overline{
Rollout 0
,5,7 have exponent 1, and exponents for 2,3,5,7
Rollout 11
Alternatively, consider the expressions for p and q and cube roots of unity. Suppose a, b, c
Rollout 11
Given abc=1, then taking reciprocals, (1/a)(1/b)(1/c)=
Rollout 2
1≡16 mod125\n\nSo now we have 16 * (-1)^{
Rollout 11
use this.\n\nAlternatively, maybe using Vieta’s formula. If we consider a polynomial equation whose roots are
Rollout 5
75) = 2/3. Cross-multiplying: 3(b + 25
Rollout 0
and \(b = 20!\), or vice versa; however, \(a\) and \(b
Rollout 0
\(b\). So for each prime factor, we assign it to either \(a\) or \(b
Rollout 11
(p) = p.\n\nGiven that complex conjugate of a sum is sum of conjugates. Similarly,
Rollout 11
x = \overline{a}, y = \overline{b}, z = \overline
Rollout 11
a, b, c. So the conjugate variables x, y, z also satisfy the same product
Rollout 0
13,17,19 are 1 (each odd?), and the others are even.

DOWN_PROJ

Top 16 Positive Activations
Rollout 9
celes triangle with base EM: if it's vertical, then the base is vertical. But triangles are
Rollout 9
triangle GEM with base EM. So EM is one of the sides of the square, the bottom side
Rollout 9
triangle. The triangle is isosceles with base EM, so the legs would be GE and GM
Rollout 9
apex of the triangle opposite the base EM, which is the vertical side of the square. If the altitude
Rollout 9
GEM is an isosceles triangle with base EM, so EM is from (10,
Rollout 9
. Since the triangle is isosceles with base EM, the apex is G. And the altitude
Rollout 9
the triangle opposite the base EM, which is the vertical side of the square. If the altitude is,
Rollout 9
base EM: if it's vertical, then the base is vertical. But triangles are usually thought of as
Rollout 13
0 * 1 = 6000\n\n6000 * 0.6 =
Rollout 9
: if it's vertical, then the base is vertical. But triangles are usually thought of as having a
Rollout 9
). But the triangle being isosceles with base EM. Since triangle GEM is isosce
Rollout 9
with base EM. So EM is one of the sides of the square, the bottom side from E to
Rollout 13
3600 = 9600, as above)\n\n6000 *
Rollout 9
have an isosceles triangle GEM with base EM. So EM is one of the sides of
Rollout 9
the left of EM. If triangle GEM has base EM (at x=10), and is
Rollout 8
ones) are adjacent and all green cards (regardless of which ones) are adjacent. Therefore, the
Top 16 Negative Activations
Rollout 11
Hmm, complex numbers with product 1. So abc = 1. None of them are real or
Rollout 1
l is non-empty if and only if r1 + r2 >= l and |r1 - r
Rollout 11
b|^2 \overline{a} →Multiply both sides by b: a = |b|^
Rollout 14
), and if it's integrable, then lim_{x→∞} f(x)=0. But
Rollout 1
that even for three points, you can find such a y. But in higher dimensions, there's more
Rollout 14
sqrt{J}} \implies A = \left( \frac{2K}{\sqrt{
Rollout 14
[f]^3 [x]^3,\n\n[J][K] = [f]^2 [x]
Rollout 11
variables? If abc =1, set variables w = ln a, z = ln b, v =
Rollout 11
For example, let me set b = a * z, c = 1/(a*b ) =
Rollout 14
K_λ=1. Then:\n\nFrom J_λ=1: λ^{2a -1
Rollout 1
2, the intersection is non-empty if r1 + r2 >= d. In our case, r
Rollout 11
, which for complex t would imply that t lies on the unit circle oris real. But since a
Rollout 11
mmm.\n\nMoreover, since abc =1, a * a z *1/(a2 z )=
Rollout 11
but since abc = 1, if |a||b||c| = 1, so if
Rollout 6
sets with few strings per length. The idea might be to encode the advice strings α_n into S_L
Rollout 2
8. Since 5 and8 are coprime, m must be ��0 mod8.

Layer 33

GATE_PROJ

Top 16 Positive Activations
Rollout 3
8,436). Let's see, 168^2 = 28,
Rollout 3
8,408). Let's see.\n\n168^2 = (170 -
Rollout 3
Alternatively, precise sqrt(26050)=161.4, as 16
Rollout 4
0*2985. Let's compute that.\n\n1990 * 298
Rollout 3
sqrt(26,049) �� 161.4 m/s. So
Rollout 3
6). Let's see, 168^2 = 28,224.
Rollout 3
33,476.2) �� 183 m/s. So that's
Rollout 3
sqrt(28,433) �� 168.6 m/s.\n\nSo
Rollout 3
50)=161.4, as 161^2=25,9
Rollout 8
0. Let's see: 240×30 =7200, 7
Rollout 3
.\n\nsqrt(28,436). Let's see, 168^2 =
Rollout 3
8,408.12) �� 168.5 m/s\n\nSo
Rollout 3
3). Let's compute with square root method:\n\n168.0^2=28,
Rollout 3
2 = 28,224. 169^2 = 28,
Rollout 3
28,436). Let's see, 168^2 = 28
Rollout 3
=28,224.0\n\n168.5^2=168
Top 16 Negative Activations
Rollout 12
(1 -k2)K(k) ] / (k(1 -k2)) * (
Rollout 12
1 -k2)K(k) ] / (k(1 -k2)) * ( -
Rollout 12
) ] / (k(1 -k2)) * ( -nu / k )\n\nSimplify:\n\n
Rollout 14
1 - μ x )dx = [1/(2λ)] ��0^{1/μ}
Rollout 12
ν2)(1 - t2) ) / (1 - ν2 ) ] / t\n\nS
Rollout 12
(1 - k2)K(k)) / (k (1 - k2)) . Let's
Rollout 12
(k) - nu2 K(k) ] / (k nu2 )\n\nThen, multiplying by dk/d
Rollout 12
(1 -k2)K(k) ] / (k(1 -k2))\n\nYes,
Rollout 12
(k) - nu2 K(k) ] / ( (1 - nu2) nu )\n\n= -
Rollout 12
(k) - nu2 K(k) ] / (k nu2 ) * (-nu /k )\n\n
Rollout 12
k2)K(k) ] / (k(1 -k2)) * ( -nu /
Rollout 12
- (1 - k2)K(k)) / (k (1 - k2)) . Let
Rollout 14
3 μ2 )\n\nSo K= [1/(2λ)] [ (1/(2 μ2 ))
Rollout 12
[ E(k) - (1 -k2)K(k) ] / (k(1 -
Rollout 12
- (1 -k2)K(k) ] / (k(1 -k2)) *
Rollout 12
(k) - nu2 K(k) ] / (k * (nu2) ) * ( -

UP_PROJ

Top 16 Positive Activations
Rollout 4
0036058 as a decimal, so yes, 0.3605
Rollout 3
)/337≈0.047. Thus, sqrt≈168.5 +
Rollout 13
1.12550881, therefore 1.12550
Rollout 13
0470643736\n\nSo 6000 * 0.
Rollout 7
9/6≈8.1667\n\nThus:\n\nA=12*(49/
Rollout 9
10 -20 = -10, left outside the square. So if h>1
Rollout 13
736 - 1.6 = 0.004706437
Rollout 12
2. Let k approach1, so x approaches0. Then K(k) ~ - (1/
Rollout 11
\n\nTherefore, cosθ = -2/5, which would allow theta = arccos(-2
Rollout 13
36 = ?\n\nCompute separately:\n\n0.07351936 * 60
Rollout 3
/337≈0.047.\n\nTherefore, sqrt≈168.5 +
Rollout 11
2/15) = -2/5\n\nTherefore, cosθ = -2/5,
Rollout 3
9,953.6 minus 66.512*2=133
Rollout 10
mod5=4; not 0 or 2, so it's a winning position.\n\n20
Rollout 7
and x must be <= R = N/3. Also, the original repeaters include x students with
Rollout 3
3.6 minus 66.512*2=133.024
Top 16 Negative Activations
Rollout 5
00 \). The midline of the trapezoid, which connects the midpoints of the
Rollout 5
\) of the segment that divides the trapezoid into two regions of equal area, we use
Rollout 6
sets \( S_1, S_2, \ldots, S_k \) by tagging each
Rollout 0
) and \(\gcd(a, b) = 1\), each prime factor of 20
Rollout 7
59 for non-promoted participants. \n\n### Part (a)\n\n1. **Original equations:
Rollout 7
= 2R \) and \( N = 3R \), so \( N \) must
Rollout 11
\left(\frac{1}{a} + \frac{1}{b} + \frac{
Rollout 6
not provide the nondeterminism inherent in \( \textbf{NP} \).\n\n**Answer:** \n
Rollout 8
\): \(2 \times (5 \times 4 \times 5 \times 4 \
Rollout 11
q) = (-1, 3)\).\n\nThe final ordered pair is \(\boxed{(-1
Rollout 1
y) \) and \( \frac{\sqrt{2}}{d}(x' - y)
Rollout 11
1, mutual multiplicative inverses), it is deduced that the only possible values that satisfy all conditions
Rollout 7
<|im_start|>answer\nGiven a mathematics test with \( N \) participants (\( N < 40 \
Rollout 13
$6,000\n- r = 12% = 0.12\n
Rollout 1
\) with radius \( \frac{d}{\sqrt{2}} \) can be non-empty
Rollout 11
{1}{b} = ac\), and \(\frac{1}{c} = ab\

DOWN_PROJ

Top 16 Positive Activations
Rollout 3
28,224.\n\n168.5^2= (168 +
Rollout 3
,224.0\n\n168.5^2=168^2 +
Rollout 5
100 = 181.25, so the greatest integer not exceeding this is
Rollout 3
is four significant digits (if we consider the trailing.5 as a justified digit). However, based on
Rollout 13
55\n---------------\nFirst, (five ten-thousandths place):\n0.00056
Rollout 12
F(φ, k) = ��_0^φ [ 1 / sqrt(1 -
Rollout 4
30,000 and 10*15=150. So total would
Rollout 4
64% ��0.8775%\n\nBut maybe doing precise sqrt(0.0
Rollout 11
zero modulo 2pi. �� + theta -2θ=0. So anyway, perhaps leading
Rollout 12
F(φ, k) = ��_0^phi [ 1 / sqrt(1 -
Rollout 14
sqrt(J) * sqrt(A )= sqrt(J)*sqrt(2K / sqrt(J))= sqrt
Rollout 11
but as in the problem there are no real root s, hence no, such cubic cannot have real coefficients
Rollout 11
to be real:\n\nImaginary part: y - y/(x2 + y2 )=0
Rollout 4
0,250. 5,940,150 -5,92
Rollout 7
impossible. Hence under any circumstances,given these constraintsthe answer to part(b) is that there is
Rollout 13
0:\n\nFirst, 0.004 of 6000 is 24.\n\n
Top 16 Negative Activations
Rollout 15
1? Wait, n=1 is 2^1 -1. Let's try n=3
Rollout 3
is a gas at standard temperature and pressure, with boiling point at -61.7 °C,
Rollout 5
bases that divides the trapezoid into two regions of equal area, its length x should be such
Rollout 15
's try n=3, which is 2^2 -1: no, since n=3
Rollout 5
the legs divides the trapezoid into two regions with areas in the ratio 2:3.
Rollout 5
bases that divides the trapezoid into two regions of equal area, the length x can be found
Rollout 15
minus one? Because n=1 is 2^1 -1 =1, and then maybe
Rollout 8
of red and one block of green (which could be in either order), or all reds or all
Rollout 9
be a large triangle with base EM (the right edge of the square) and apex G outside the square
Rollout 9
10 -100/h)*x ] from 0 to10.\n\nEvaluating at
Rollout 15
Wait, so OO maps to OO. That's a problem.\n\nWait, if we start with OO at
Rollout 5
midline divides the trapezoid into two regions with areas in the ratio 2:3.
Rollout 15
and first subdiagonal.\n\nWait, example for n=4:\n\nM would be:\n\n0 1
Rollout 9
Wait, no. Wait, EM is the top side of the square. Wait, hold on. \n\n
Rollout 9
with base EM. So EM is one of the sides of the square, the bottom side from E to
Rollout 15
, they're still OO. Therefore, this is a fixed point. Thus, the system doesn't die

Layer 34

GATE_PROJ

Top 16 Positive Activations
Rollout 9
is 100 -500/h=80. Then 500/h=
Rollout 9
(10,0):\n\nUsing point-slope form: y -0 = (-5/h)(x
Rollout 7
P = 56N +15P=66N\n\nSo 15P=
Rollout 11
3 - sx^2 + tx - 1 = 0, where s = a + b +
Rollout 14
λ μ2=1/12\n\nDivide (1) by (2):\n\n(λ2
Rollout 11
, combining p and q into a system of equations. For example, ifq is something like s^
Rollout 11
expands to something that can be combined with q, but seems messy.\n\nAlternatively, writing in terms of recip
Rollout 3
standard temperature to assume. Let me check online, but since I can't do that, I need to
Rollout 11
where p and q are real coefficients and equations. But without real roots. But from Descarte’s signs
Rollout 11
t, and the constant term is -1.\n\nBut given that a, b, c are complex numbers
Rollout 15
polynomials.\n\nIn paper titled "Nilpotent\n\n**Final Answer**\nAll positive integers that are powers
Rollout 11
c, t=ab + bc + ca, but here p =s +t. Hmm. The
Rollout 14
Lagrangian multipliers given the constraints.\n\nLet’s think in terms of optimization: maximize I=∫
Rollout 11
Alternatively, maybe take conjugate of variables. Let’s set x = \overline{a}, y
Rollout 14
)= (1 -mu x)/(2 lambda ). But then in such case, it gave a not very
Rollout 5
up the equation: (b + 25)/(b + 75) = 2/
Top 16 Negative Activations
Rollout 2
^1 -1), the second is k=2 (10^2 -1), ..., up
Rollout 2
1)=9 mod125\n\nk=2:10^2 -1=99
Rollout 10
maybe in each two steps, the losing positions are even, odd, even, odd. Wait, starting
Rollout 2
2 -1 =1 mod8\n\nk=2: (4 -1)=3 mod8\n\n
Rollout 10
→Grundy[6}=1\n\nn=7: subtract1→6 (Grundy1);
Rollout 10
5 (L). Therefore, W.\n\nn=7: n -1=6 (W); n
Rollout 2
modulus125, each term from k=3 onward is≡-1, so multiplying 9
Rollout 10
Therefore, n=2 is L.\n\nn=3: n -1=2 (L). So
Rollout 10
=3 (W). So L.\n\nn=8: n -1=7 (L). So
Rollout 2
:10 -1=9.\n\nk=2:100 -1=99.\n\n
Rollout 2
00 -1=99.\n\nk=3:1000 -1=99
Rollout 10
large k, the difference between k2 and (k+1)2 is 2k+1
Rollout 10
between consecutive squares (k2 and (k+1)2), the possible moves that can be made
Rollout 2
first two terms (k=1 andk=2) are9,99. The remaining
Rollout 10
step:\n\nn=0: L\n\nn=1: Check n - 1=0. Since
Rollout 2
=99 mod125\n\nk=3:10^3 -1=99

UP_PROJ

Top 16 Positive Activations
Rollout 8
original problem, seems that my answer is right.\n\nTherefore, probability is 31/126
Rollout 8
800 = 7200\n\nTherefore, total happy sequences:\n\nCase1 + Case2
Rollout 2
methods confirm that the remainder is109.\n\nTherefore, answer:109.\n\nBut let me
Rollout 3
0.222) )\n\nFirst compute numerator:\n\n8*8.314 = 6
Rollout 13
difference is $187.12.\n\nTherefore, the answer is $187.1
Rollout 2
91≡16 mod125\n\nSo now we have 16 * (-1)^
Rollout 12
\n\nSecond term: (1 - t2)\n\nSo denominator becomes sqrt( ( (t2 - ν
Rollout 2
5=109 mod125.\n\nTherefore, P ��109 mod12
Rollout 0
19. So that's 8 primes. Therefore, the number of coprime pairs would be
Rollout 10
, 6 is winning, 3 is winning. So 7 cannot move to a losing position.
Rollout 3
. So approx 161 m/s.\n\nSo depending on the temperature assumption, the answer would be
Rollout 2
) = -3 ��5 mod8.\n\nTherefore, P �� 5 mod8.\n\nAlright
Rollout 5
25y\n\nExpanding the left-hand side:\n\ny^2 - 125y +
Rollout 4
So approximately 0.8776%.\n\nThus, the percentage error is approximately 0.8
Rollout 13
441.11616\n\nTherefore, A_annual �� $9,4
Rollout 3
figs justify 169 m/s.\n\nTherefore the valid boxed answer is \boxed{16
Top 16 Negative Activations
Rollout 14
*1? Not sure either.\n\nAlternatively, use homogeneous functions. Suppose f is a scaling function such that
Rollout 8
available.\n\nTherefore, this is similar to permutation with two colors. The number of such sequences is C(
Rollout 1
orthogonal to the set S - S, except for a constant.\n\nBut in that case, if we take
Rollout 1
with equal norms and their differences have equal norms, this often relates to them being vertices of a simplex or
Rollout 15
lamps will eventually turn off after some time.\n\nFirst, let me rephrase the rules to make sure I
Rollout 1
of them.\n\nAlternatively, for the system to have a solution, then for any x, x' in
Rollout 1
them as orthogonal vectors. The proof existence uses the separability and completeness.\n\nBut maybe an innovative approach is
Rollout 1
2, for all x in S, has a common solution y in H. Alternatively, the intersection
Rollout 14
(x). Hmm.\n\nAlternatively, recognize that via Homogeneity: The inequality is scaling invariant. If we replace
Rollout 1
- y|| = d / sqrt(2) } is not convex. Oh, right. The sphere
Rollout 14
that. Suppose f_λ(x) = λ^{a} f(λ x). Then:\n\nCompute
Rollout 14
infty f(x)^2 dx \right) \left(\int_0^\infty xf(x
Rollout 1
and their differences have equal norms, this often relates to them being vertices of a simplex or something similar.
Rollout 1
intersection is non-empty. So, since each sphere is closed and convex, and in a Hilbert space
Rollout 14
dimensionally consistent. Therefore, the inequality is scaling-invariant. Therefore, without loss of generality,
Rollout 14
) is of the form $f(x) = c/(1 + d x)$, but that's

DOWN_PROJ

Top 16 Positive Activations
Rollout 14
( \int f(x)^2 dx - C \right) - \mu \left( \int
Rollout 14
integral:\n\n∫ x dx = x2 /2 from 0 to 1/μ: (1
Rollout 9
10 -100/h)*x ] from 0 to10.\n\nEvaluating at
Rollout 9
x^2/2) +6x ] from0 to10\n\n= [ (1/
Rollout 14
2 dx = μ * x^3 /3 from 0 to1/μ: μ*(1
Rollout 9
the altitude to EM is the perpendicular distance from G to EM. Since EM is the vertical line from (
Rollout 14
function for f is different. When I applied Lagrange multipliers for f(x) with two constraints,
Rollout 9
, as I thought, the horizontal distance from G to EM (the line x=10), then
Rollout 14
dx -λ ��f2 dx - μ ��x f dx.\n\nTaking the functional derivative,
Rollout 9
= (5/h)x - (50/h) +10.\n\nWhich is y = (5
Rollout 9
But wait, if h is the altitude from G to EM, h can be anything. Let's think
Rollout 14
μ ) - μ*(x2 /2 ) from0 to1/μ ]= [1/(
Rollout 3
to message user to specify temperature, but as IA cannot, so need to make a call. \n\n Since
Rollout 9
, then if h is the distance from G to EM, then if G is inside the square, h
Rollout 2
1 mod125: 125*7=875. 891
Rollout 9
= (-5/h)x + (50/h) +0? Wait, let's verify:\n\nFrom
Top 16 Negative Activations
Rollout 14
three functions f, g, h, and exponents p, q, r with 1/p +
Rollout 14
lder's inequality for three functions would state that for exponents $p, q, r$ such that
Rollout 14
dx.\n\nThen apply Hölder's inequality with exponents p, q, r such that 1/p
Rollout 14
Let’s remember that Hölder’s inequality with exponents 2 and 2 gives Cauchy-S
Rollout 14
} dx.\n\nThen apply Hölder's inequality with exponents p, q, r such that 1
Rollout 14
3/2? Wait, Hölder's with exponents p, q such that 1/p +
Rollout 14
.\n\nLet’s remember that Hölder’s inequality with exponents 2 and 2 gives Cauchy
Rollout 14
's inequality for three functions would state that for exponents $p, q, r$ such that $
Rollout 14
/2} dx.\n\nApply Hölder with exponents p=2, q=4, and r
Rollout 14
? Since Hölder can relate multiple integrals with exponents that are conjugate. Let me recall Hö
Rollout 14
Hölder's inequality with exponents 3, 3, and 3/2? Wait,
Rollout 14
2a -1)) ln J = (1/(a -2)) ln K.\n\nCross-multip
Rollout 14
lder's inequality: for $p$ and $q$ such that $1/p + 1/q
Rollout 14
Alternative idea: Use Hölder's inequality with exponents p =2 and q=2, wait.\n\n
Rollout 14
dx?\n\nAlternatively, Hölder's inequality with exponents 3, 3/2, and
Rollout 14
ölder's inequality for three functions would state that for exponents $p, q, r$ such

Layer 35

GATE_PROJ

Top 16 Positive Activations
Rollout 0
answer is 128, corresponding to \boxed{128}? Or wait, wait
Rollout 4
2 + (2 * 0.004)^2 )\n\nCalculating 2*0.
Rollout 4
90=4975*(1200-10)=4975*
Rollout 3
the mean molecular speed of radon gas is \boxed{170} m/s if we round
Rollout 9
answer is 25. Therefore box form \boxed{25}\n\nWait, that seems correct.\n\n
Rollout 4
error in the calculation of the energy stored is \boxed{0.88\%}.\n\n<|im_start|>
Rollout 3
K (unless stated otherwise), perhaps answer around \boxed{169} m/s. But need
Rollout 12
(1 - ν2 - t2 + ν2)/(1 - ν2) ) = (1
Rollout 0
<=N). For example, primes between N/2 and N cannot come in pairs. So hence ex
Rollout 11
1,3). Hence, ordered pair is \boxed{(-1, 3)}.\n\n**Final
Rollout 3
m/s. Thus, present answer boxed as \boxed{169 \text{ m/s}}
Rollout 11
, 3)}.\n\n**Final Answer**\n\boxed{(-1, 3)}\n\n<|im_start|>answer\n
Rollout 4
that.\n\n1990 * 2985= (2000 -10
Rollout 5
The midline length is (b + (b + 100))/2 = (2b
Rollout 2
109.\n\nTherefore, boxed answer is \boxed{109}\n\n**Final Answer**\n\
Rollout 3
19,944 – 132.96=19,811
Top 16 Negative Activations
Rollout 14
1 - μ x)/(2 λ)] dx.\n\nSo first, let's compute J:\n\nLet’s do
Rollout 4
\n\nHmm, let's compute these step by step.\n\nFirst, compute the coefficients for ΔC_eq.\n\n
Rollout 14
λ), and zero elsewhere.\n\nCompute the integrals:\n\nJ = ��0^{x_max} [(
Rollout 13
to multiply directly. Still, this might take time.\n\nAlternatively, using currency calculation steps:\n\n1.5
Rollout 1
||y||2 -d2/2.\n\nTherefore, each inner product of xi with y is this
Rollout 2
need to compute P mod 1000.\n\nNow, let's approach this modulo 8 and
Rollout 1
inner product of xi with y is this value.\n\nSo, assuming that all such equations hold.\n\nBut if
Rollout 8
compute this for k=1 to k=4.\n\nk=1:\n\n2 * P(5,
Rollout 1
⟩ - ⟨x, x⟩ = k.\n\nTherefore, 2⟨x, y⟩ = ⟨
Rollout 1
, we might use this to relate their norms.\n\nWait, but unless all the points in S have the
Rollout 2
1 to 999 mod 8.\n\nHmm. Let's compute this product. But before multiplying
Rollout 11
/4) e^{-i2θ}.\n\nThus, p = 2*(2 e^{i
Rollout 14
Therefore, determining λ and μ to satisfy the constraints.\n\nSo suppose x ranges up to x_max=1
Rollout 7
equation with x and N. Let's compute that.\n\nFirst expand:\n\n75*(2N/3
Rollout 15
: all on (OOOO), let's see:\n\nt=0: O O O O (L
Rollout 6
), but this is still not polynomial per length. Wait.\n\nAlternatively, let's encode the entire α_n

UP_PROJ

Top 16 Positive Activations
Rollout 7
?\n\nWait, wait:\n\nWait 18N is equal to 54N/3. So
Rollout 1
2 = 3d2/4 - which is problematic.\n\nHowever, at the same time,
Rollout 6
are in $\textbf{P}_{angel}$ because the \textit{angel string}
Rollout 11
(1 + a)(1 + b)(1 + c) = 1 + (a + b
Rollout 11
is prohibited. Hence, r3 +1/r3 ≥2, which would mean cosθ ≥1
Rollout 11
+ bc + ca) + 2. Wait, so then p = (1 + a)(1
Rollout 1
2 = 3d2/4 - which is problematic.\n\nHowever, at the same time, the
Rollout 11
, which is prohibited. Hence, r3 +1/r3 ≥2, which would mean cosθ
Rollout 14
8)/(4/9)=9/(32)= approximately 0.28125 <
Rollout 7
1R = 66N +5N =71N.\n\nBut separately, we can say
Rollout 7
. Which is 56*4=224. Correct.\n\nAfter increase, promoted total score
Rollout 14
44 /64= 9/4=2.25.\n\nThen 8 J K
Rollout 11
(1 +a)(1 +b)(1 +c)=2.\n\nBut any way, not directly
Rollout 12
E(k) - nu2 K(k) ] =E(k) - (1 -k2)
Rollout 7
, wait:\n\nWait 18N is equal to 54N/3. So 5
Rollout 14
]^3 [x]^3,\n\n[J][K] = [f]^2 [x] * [
Top 16 Negative Activations
Rollout 5
+ 175)(125H)/(75 + x) = 125
Rollout 5
125(x + 175)]/(x + 75) = 12
Rollout 9
. However, an isosceles triangle with base EM: if it's vertical, then the base
Rollout 0
a}{b}\) where \(0 < a < b\) and \(a\) and \(b\)
Rollout 9
triangle. The triangle is isosceles with base EM, so the legs would be GE and GM
Rollout 0
\(a\) and \(b\) are coprime positive integers. The product \(a \times b
Rollout 5
125(x + 175)/(x + 75)\n\nLet me write the
Rollout 0
= 20!\), \(0 < a < b\), and \(\gcd(a, b
Rollout 0
b = 20!\) and \(a < b\).\n\nNow, how do I count such
Rollout 9
, the area of overlap between triangle GEM and square AIME is 80, which is
Rollout 0
20!\), \(0 < a < b\), and \(\gcd(a, b) =
Rollout 9
and M are on the square, and G is inside, so triangle is within. But then how come
Rollout 0
20! where each pair consists of two coprime numbers.\n\nBut since the fraction is between
Rollout 0
20!. So that equivalent to saying find all reduced fractions a/b where 0 < a/b <
Rollout 14
,\n\n$K = a \int_0^b x dx = a (b^2)/2
Rollout 0
such coprime pair where \(a < b\), we have a distinct rational number, and each

DOWN_PROJ

Top 16 Positive Activations
Rollout 3
=8.31 instead of 8.314. Let me check. Suppose R=
Rollout 7
score must be 98/1=98, but >=60 and <65.
Rollout 7
8. But the original scores can’t be98, since they need to be between60-
Rollout 11
the function f(r) =r3 +1/r3.\n\nIf r >1, then f(r
Rollout 2
9*1000=9999-9000=999.
Rollout 9
from (0,0) to (0,10)). Since G is to the left of the
Rollout 2
mod1000= 9999-9*1000=99
Rollout 2
99 \times \cdots \times \underbrace{99\cdots9}_{\
Rollout 4
2.774 * 3.1623e-3 �� 8
Rollout 2
*37.\n\nFourth term:9999=999*10 +9,
Rollout 3
0 +70 +3)*66.512 =\n\n200*66.
Rollout 3
K=273 with R=8.314:\n\nWe did 161 m
Rollout 2
+9, wait, perhaps 9999=9*1111, 1
Rollout 3
. Let me check. Suppose R=8.31:\n\n8*8.31=6
Rollout 3
2*298 ��24.942*300=7,4
Rollout 3
\n\nSum:13,302.4 +5,986.08=
Top 16 Negative Activations
Rollout 7
3P=2N => P= (2/3)N.\n\nYes that's correct. So
Rollout 8
5−r} or G^{5−r} R^r. So two options.\n\nIn terms
Rollout 3
25,921;162^2=26,244. So
Rollout 11
have | | >1, or all | | <1, or some mix. But their product is
Rollout 5
, we need to find x, which is the length of the segment that divides the trapezoid
Rollout 7
.\n\nSimilarly, the remaining repeaters are (N/3 -x) = (N/3 -
Rollout 14
is $I^3$, and the right-hand side is $8$ times $J K$. So
Rollout 8
of greens, either reds first then greens or greens first then reds. For each possible split of
Rollout 7
3P=2N => P = (2/3)N.\n\nSo N must be divisible by
Rollout 8
red block then green block, or first the green block then red block. Therefore, for k =
Rollout 8
2, same for all green. For the mixed color sequences (either RG or GR):\n\nThey require that
Rollout 5
we need to find x, the length of the segment that divides the trapezoid into two regions
Rollout 5
of equal area. \n\nFor a line segment parallel to the bases that divides the trapezoid into
Rollout 0
is between 0 and 1, \(a < b\). So for each such coprime
Rollout 7
with P + R:\n\n71P + 56R = 66(P + R)\n\n
Rollout 4
10 / 2000 = 0.005. Similarly, for C

Layer 36

GATE_PROJ

Top 16 Positive Activations
Rollout 14
Therefore, determining λ and μ to satisfy the constraints.\n\nSo suppose x ranges up to x_max=1
Rollout 8
)! ) * (5! /k! ) ]\n\nThen, probability is [240 + Sum
Rollout 5
, let me check the calculations again step by step. Probably, I made an algebraic mistake.\n\nStarting
Rollout 7
't be. So there must be a mistake here. Let me verify step by step.\n\nWe derived:\n\n
Rollout 9
) with base EM.\n\nSo let's calculate coordinates.\n\nGiven:\n\nE is (10,0)\n
Rollout 8
30240.\n\nSimplify this fraction.\n\nFirst, divide numerator and denominator by 24
Rollout 7
equation with x and N. Let's compute that.\n\nFirst expand:\n\n75*(2N/3
Rollout 13
51936\n\nCalculating that value:\n\nFirst, 1 * 6000
Rollout 1
achieve this for all x, x' in S?\n\nSuppose that all points in S lie on a
Rollout 0
instead of 20!, take a simpler case. Let me check with n = 6. Then
Rollout 7
need to model the original groups and the new groups.\n\nLet me denote variables:\n\nOriginal scenario:\n\n- Number
Rollout 4
\n\nHmm, let's compute these step by step.\n\nFirst, compute the coefficients for ΔC_eq.\n\n
Rollout 0
the exponents of primes in 20!:\n\nFor prime 2: floor(20/
Rollout 1
k such that this holds for all x in S.\n\nBut maybe we can see this as a functional equation
Rollout 8
0P5\n\nAlternatively, compute numerator and denominator.\n\nFirst, compute denominator: 10P5
Rollout 3
22)) Hmm, okay. Let me calculate.\n\nFirst, 8 * 8.31
Top 16 Negative Activations
Rollout 6
's theorem equivalence between \( \textbf{P}/\text{poly} \) and sparse or
Rollout 6
}$ . Show that there exists a sparse set $S_L$ and a deterministic polynomial time TM
Rollout 11
a} = |b|2 a \overline{b} → Then original equation: a =
Rollout 14
\sqrt{A} + \frac{K}{A} \), we solve for \( A \
Rollout 15
if and only if at minute $t$ , exactly one of its neighbors is on (the
Rollout 14
\sqrt{A}} - \frac{K}{A^2} = 0 \implies
Rollout 3
68*0.5 +0.5^2=28,224 +1
Rollout 6
bb{N}$ , a sequence of strings $\{\alpha_n\}_{n \in \math
Rollout 14
1 with1/p +1/q=1,\n\nab≤ (a^p)/p +(b^
Rollout 6
theorem equivalence between \( \textbf{P}/\text{poly} \) and sparse oracles
Rollout 5
left-hand side:\n\ny^2 - 125y + 100y -
Rollout 3
168*0.5 +0.5^2=28,224 +
Rollout 11
a} = |b|^2 a \overline{b}\n\nMultiply both sides by b:\n\n\
Rollout 6
,1\}^{p(n)}$ , and a deterministic polynomial time Turing Machine $M$
Rollout 11
(b c)/a^2 + c^2/(a b) ]\n\nHence, q *
Rollout 5
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nOne base of a trapez

UP_PROJ

Top 16 Positive Activations
Rollout 6
doesn't work. Another idea is to use pairwise hash functions or some indexing scheme. Wait, but we
Rollout 2
1000 quickly converges due to multiplicative factors. So the answer is109.
Rollout 0
20!.\n\nTo compute the exponents of primes in 20!:\n\nFor prime 2:
Rollout 0
in N's prime factorization), which requires multiplicativity. For each prime p | N, assign all
Rollout 0
and 144. How is it multiplicative, related to the number of prime factors?\n\nYes
Rollout 6
't work. Another idea is to use pairwise hash functions or some indexing scheme. Wait, but we need
Rollout 11
1, the constant term would be -1 (for cubic x^3 - sx^2 + tx
Rollout 5
\) and the longer base be \( b + 100 \). The midline of the
Rollout 0
exponents), it depends. As with N! for N>1.\n\nWait, for N >=2
Rollout 12
k), where sn is the Jacobi elliptic function. Wait, maybe not necessary. Alternatively, use
Rollout 14
step:\n\nsqrt(A)=sqrt(2K/sqrt( J ))}= (2K/ J^{
Rollout 7
with \( N \) participants (\( N < 40 \)), the pass mark is fixed at
Rollout 2
when you have multiple terms multiplied, perhaps with some factors of 1000. If the product
Rollout 2
\nTo find the remainder when \(9 \times 99 \times 999 \times
Rollout 6
1/k}, assuming m is a perfect kth power. Otherwise, no solution. Hence, for each
Rollout 13
2}{1}\right)^{1 \cdot 4} = 6000 \left
Top 16 Negative Activations
Rollout 9
E=(10,0), M=(10,10), G=(10 - h
Rollout 9
0,10), and E at (10,0). That's consistent.\n\nNow triangle G
Rollout 9
(10,0)\nM is (10,10)\nG is (10 -
Rollout 9
5), (10,0), (10,10). The overlapping area between this triangle
Rollout 9
E (10,0), M (10,10). So the altitude is indeed h
Rollout 9
coordinates:\n\nA(0,0), I(0,10), M(10,1
Rollout 9
to (10,0) to (10,10). The area of this triangle can
Rollout 9
triangle spans from (10 - h,5) to (10,0) and (1
Rollout 9
EM is the vertical line from (10,0) to (10,10), the
Rollout 9
(10,10), E(10,0). Wait, wait no: in the
Rollout 8
0, all green, and k=5, all red, which are already covered in Case 1
Rollout 9
A at (0,0), I at (0,10), M at (10,
Rollout 14
2 = [f]^3 [x]^3.\n\nSo dimensionally consistent. Therefore, the inequality is
Rollout 10
(losing). Therefore, n=1 is winning.\n\nn=2: from 2, you can
Rollout 1
||y||2 = d2 / 2.\n\nThus:\n\n-2⟨x, y〉 +
Rollout 10
Let me write them down:\n\n0: losing\n\n1: win (can take 1, leaving

DOWN_PROJ

Top 16 Positive Activations
Rollout 9
= 100 => y = 5.\n\nTherefore, the y-coordinate of point G must be
Rollout 14
int_0^\infty f(x)^2 dx$,\n\n$K = \int_0^\
Rollout 8
probability that Kathy will be happy is $\frac{m}{n}$ , where $m$ and $
Rollout 14
1})^{1/(2a -1)}}\n\nFrom K_λ=1: λ^{a
Rollout 3
·mol). Then, dividing by M in kg/mol, the mol unit cancels, then kg in
Rollout 14
int_0^\infty f(x)^2 dx \right) \left(\int_0^\
Rollout 14
, if you use 1, which is not integrable unless we apply some weight.\n\nAlternatively consider using
Rollout 2
1000=2^3 *5^3.\n\nTherefore, if in the product there
Rollout 7
/3.\n\nBut x= N/12.\n\nSo substituting:\n\n(N/12)(A
Rollout 4
, their product is 1990*2985. Let's compute that.\n\n1
Rollout 6
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet's say a language $
Rollout 7
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nIn a mathematics test number of participants
Rollout 10
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nConsider the following two person game.
Rollout 8
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nKathy has $5$ red
Rollout 9
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nSquare $AIME$ has sides
Rollout 11
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nSuppose $a,\,
Top 16 Negative Activations
Rollout 12
if that helps. Alternatively, substitution s = t2 - ν2?\n\nAlternatively, substitution φ where t
Rollout 1
||2. Therefore:\n\nd2 = (d2)/2 + (d2)/2 -
Rollout 3
2)^2 = 170^2 - 4*170 + 4 =
Rollout 6
, but here the advice is fixed for all inputs of length n, and the machine runs in poly-time
Rollout 8
actually, for k=2, is the number 1200 per sequence? Wait, no
Rollout 1
circular.\n\nAlternatively, setting t = ||x|| squared, you get equations:\n\n-⟨x, y
Rollout 6
allows L to be reduced. So probably the intended answer is to cite/to use this theorem.\n\nHence
Rollout 11
, perhaps take a, b, c to be complex numbers lying on some specific curves? Not sure.\n\n
Rollout 10
a non-periodic structure.\n\nStarting from n=0 (L), n=1 (W), n
Rollout 8
10! / (10-5)! = 10 * 9 * 8
Rollout 3
is wrong, but maybe the problem's answer key uses 298 K.\n\nAlternatively, let's
Rollout 2
8, and 2^k for k ≥3 is 0 mod8. Thus, for
Rollout 11
a)(1 + b)(1 + c) - 2, as established earlier. But I don
Rollout 12
} ] dt\n\nPerhaps perform substitution s = t2. Let me try:\n\nLet s = t2
Rollout 3
.5)^2=168^2 + 2*168*0.5
Rollout 10
and W represents winning positions.\n\nInitialize:\n\nn=0: L\n\nFor n >=1:\n\nn is

Layer 37

GATE_PROJ

Top 16 Positive Activations
Rollout 5
5). Is that right?\n\nWait, let me check the calculations again step by step. Probably, I
Rollout 13
a dollar value to the nearest cent.\n<|im_start|>assistant\n<|im_start|>think\nOkay, so Joanie took a
Rollout 13
.12^4. Let me do this step by step.\n1.12^1 =
Rollout 13
1.12^4. Let me do this step by step.\n1.12^
Rollout 12
$ . [P. Turan]\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to prove
Rollout 14
(x) dx \right) $$ \n<|im_start|>assistant\n<|im_start|>think\nOkay, so I have to prove
Rollout 10
win no matter how his opponent plays.\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to prove
Rollout 3
so I need to be careful here. Let me check.\n\nI recall the root mean square (rms)
Rollout 8
. Find $m + n$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. We
Rollout 13
compute 1.12^4. Let me do this step by step.\n1.12
Rollout 5
iscalculation in the algebra step. Let me verify quickly. \n\nAlternatively, perhaps it is quicker to
Rollout 13
1.12^4. Let me do this step by step.\n1.12^1
Rollout 3
(Rn) in m/s\n<|im_start|>assistant\n<|im_start|>think\nOkay, I need to calculate the
Rollout 4
15 )^2\n\nHmm, let's compute these step by step.\n\nFirst, compute the coefficients
Rollout 7
equation with x and N. Let's compute that.\n\nFirst expand:\n\n75*(2N/3
Rollout 2
by $1000$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I have this problem
Top 16 Negative Activations
Rollout 12
ν}^1 ( (t2 - ν2)^{-1/2} (1 - t
Rollout 12
expression: E(k) - (1 -k2) K(k). If this is approaching1 -
Rollout 12
ν}^1 [ (t2 - ν2)^{-1/2} (1 - t
Rollout 12
1 / sqrt( (t2 - ν^2)(1 - t^2) ) ] dt
Rollout 12
2) ln x, so (1 -k2)K(k)~ x*(- (1/
Rollout 12
{1}{\sqrt{(t2 - ν2)(1 - t2)}} dt\n\nLet me
Rollout 10
winning position. A possible set of losing positions are those numbers in the form where their base-5 decomposition
Rollout 12
E(k) approaches1 and (1 -k2) K(k) approaches0. Thus, E
Rollout 12
1 / sqrt( (t2 - ν2)(1 - t2) ) ] dt\n\n
Rollout 12
�_{ν}^1 [ (t2 - ν2)^{-1/2} (1
Rollout 12
ν}^1 [ (t2 - ν2)^{-1/2} (1 - t
Rollout 12
ν}^1 [ (t2 - ν2)^{-1/2} (1 - t
Rollout 12
K(k) ] / ( (1 -nu2)nu )= - [positive] / [
Rollout 12
{E(k) - (1 - k^2)K(k)}{k(1 - k
Rollout 12
) remains about1, whereas (1 -k2) K(k) approaches0 * infty,
Rollout 12
E(k) / (k(1 -k2)) ) ] - [ K(k) /k

UP_PROJ

Top 16 Positive Activations
Rollout 5
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nOne base of a trapez
Rollout 2
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nFind the remainder when $9 \
Rollout 13
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nJoanie takes a $\$6
Rollout 0
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nGiven a rational number, write it
Rollout 3
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nCompute the mean molecular speed v in
Rollout 1
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet $ \mathcal{
Rollout 9
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nSquare $AIME$ has sides
Rollout 7
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nIn a mathematics test number of participants
Rollout 4
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nTwo capacitors with capacitance values
Rollout 8
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nKathy has $5$ red
Rollout 14
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nA function $f:[0
Rollout 15
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet $n\geq
Rollout 10
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nConsider the following two person game.
Rollout 11
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nSuppose $a,\,
Rollout 6
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet's say a language $
Rollout 12
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nProve that the function \[
Top 16 Negative Activations
Rollout 9
be equal. Therefore, the triangle has two equal sides (GE and GM) with base EM.\n\nSo
Rollout 11
equal to the sum of the variables plus the sum of their products two at a time. That form makes
Rollout 11
p is equal to the sum of the variables plus the sum of their products two at a time. That
Rollout 11
b)(1 + c) = 1 + (a + b + c) + (ab +
Rollout 15
ent over GF(2). A matrix is nilpotent if some power of it is the zero matrix
Rollout 4
C1 + C2)) ]\n\nTherefore, the relative error in C_eq is ΔC_eq / C
Rollout 12
/2) ln x, so (1 -k2)K(k)~ x*(- (1
Rollout 5
, the ratio of the areas depends on the squares of the lengths involved. The formula for the length of
Rollout 11
variables plus the sum of their products two at a time. That form makes me think of the expansion of
Rollout 11
3θ).\n\nIt would be 1 +8(cos3θ + i sin3θ) + (
Rollout 15
). A matrix is nilpotent if some power of it is the zero matrix.\n\nSo, for M
Rollout 11
1 + (a + b + c) + (ab + bc + ca) + abc.
Rollout 15
to summing mod2. Because when the sum is exactly 1, the XOR is1, else
Rollout 15
're working over finite fields, a matrix is nilpotent if and only if all its eigenvalues (
Rollout 11
the sum of the variables plus the sum of their products two at a time. That form makes me think
Rollout 11
+ b + c, t = ab + bc + ca). Then p = s + t, which

DOWN_PROJ

Top 16 Positive Activations
Rollout 12
+ [ nu2 K(k) ] / ( (1 - nu2) nu )\n\nSimplify terms
Rollout 12
) + [ nu2 K(k) ] / ( (1 - nu2) nu )\n\nSimplify
Rollout 2
8.\n\nTherefore, equation becomes:\n\n(5m +5)≡5 mod8\n\nSubtract
Rollout 6
textbf{NP} \).\n\n**Answer:** \n\( \textbf{P} = \text
Rollout 11
^{iθ}, c=1/(r2 e^{i2θ}). So a = b =
Rollout 1
necessarily?\n\nWait, the problem doesn't state anything on how big the set S is. Given that in
Rollout 1
0.\n\nSo another way to think of this is that the set {x - y : x ∈
Rollout 4
approximately 0.88%.\n\n**Final Answer**\nThe percentage error in the calculation of the energy stored
Rollout 12
K(k) - E(k)/nu ] / (1 - nu2 )\n\nBut since k = sqrt
Rollout 9
0 -50/25=(1/5)x +10 -2= (1/
Rollout 6
\) is unique per length.\n\n**Answer:** \nThe sparse set \( S_L \) contains encoded
Rollout 12
2 - ν2)/(1 - ν2) ) times sqrt(1 - (t2 - ν
Rollout 6
\textbf{NP} \).\n\n**Answer:** \n\( \textbf{P} = \
Rollout 4
respect to a is (b*(a + b) - a*b)/(a + b)^2)
Rollout 4
.8412\n\nThus E_max≈15,195.84e-
Rollout 12
) nu )\n\nSecond term: [ nu2 / ( (1 - nu2) nu ) ] K
Top 16 Negative Activations
Rollout 5
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nOne base of a trapez
Rollout 2
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nFind the remainder when $9 \
Rollout 13
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nJoanie takes a $\$6
Rollout 0
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nGiven a rational number, write it
Rollout 4
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nTwo capacitors with capacitance values
Rollout 1
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet $ \mathcal{
Rollout 9
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nSquare $AIME$ has sides
Rollout 7
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nIn a mathematics test number of participants
Rollout 3
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nCompute the mean molecular speed v in
Rollout 6
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet's say a language $
Rollout 15
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet $n\geq
Rollout 10
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nConsider the following two person game.
Rollout 8
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nKathy has $5$ red
Rollout 12
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nProve that the function \[
Rollout 11
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nSuppose $a,\,
Rollout 14
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nA function $f:[0

Layer 38

GATE_PROJ

Top 16 Positive Activations
Rollout 5
81 is the correct answer. \n\n**Final Answer**\n\boxed{181}\n\n<|im_start|>answer
Rollout 11
{(-1, 3)}.\n\n**Final Answer**\n\boxed{(-1, 3)}\n\n
Rollout 15
Let me look for the 0\n\n**Final Answer**\nThe values of \( n \) that are
Rollout 0
8-1}=128.\n\n**Final Answer**\n\boxed{128}\n\n<|im_start|>answer
Rollout 15
In paper titled "Nilpotent\n\n**Final Answer**\nAll positive integers that are powers of two.
Rollout 1
}}" or as per problem statement.\n\n**[Final Answer]**\n\boxed{\text{Such a point
Rollout 3
{169} m/s.\n\n**Final Answer**\nThe mean molecular speed of radon gas is
Rollout 13
$187.12.\n\n**Final Answer**\n\boxed{187.12
Rollout 9
being h=25.\n\nPerfect.\n\n**Final Answer**\n\boxed{25}\n\n<|im_start|>answer\n
Rollout 15
ynomials.\n\nIn paper titled "Nilpotent\n\n**Final Answer**\nAll positive integers that are powers of
Rollout 15
paper titled "Nilpotent\n\n**Final Answer**\nAll positive integers that are powers of two. So
Rollout 8
26= 157.\n\n**Final Answer**\n\boxed{157}\n\n<|im_start|>answer
Rollout 14
with appropriately chosen vectors / weighet\n\n**Final Answer**\n\boxed{ \left(\int_0
Rollout 3
{169} m/s.\n\n**Final Answer**\nThe mean molecular speed of radon gas is
Rollout 2
is \boxed{109}\n\n**Final Answer**\n\boxed{109}\n\n<|im_start|>answer
Rollout 4
is approximately 0.88%.\n\n**Final Answer**\nThe percentage error in the calculation of the energy
Top 16 Negative Activations
Rollout 12
f(\nu)= \int_1^{\frac{1}{\nu}} \frac{dx
Rollout 4
{eq}} \) is:\n\n\[\n\frac{\Delta C_{\text{eq}}}{
Rollout 12
1 -k2) K(k) approaches0 * infty, which is more nuanced. Hmmm
Rollout 3
assistant.\n<|im_start|>user\nCompute the mean molecular speed v in the heavy gas radon (Rn)
Rollout 0
problem.\n\nFirst, a rational number between 0 and 1 can be written as a fraction \(\
Rollout 12
the integral becomes∫_{theta'=pi/2}^0 [ -d theta' / sqrt(
Rollout 4
the relative error in C_eq is ΔC_eq / C_eq = sqrt( [ (C2 /
Rollout 0
it says that the original rational number between 0 and 1 is written in reduced form, and you
Rollout 12
=ν2; t=1, s=1. Therefore,\n\nf(ν) = ��
Rollout 0
<|im_start|>answer\nGiven a rational number between 0 and 1, we need to write it as a
Rollout 12
, the integral becomes∫_{theta'=pi/2}^0 [ -d theta' / sqrt
Rollout 12
^2. Let k approach1, so x approaches0. Then K(k) ~ - (1
Rollout 4
frac{\Delta C_{\text{eq}}}{C_{\text{eq}}} = \frac
Rollout 12
f'(\nu) \) is positive, and the denominator is positive as well, leading to:\n\n
Rollout 0
First, a rational number between 0 and 1 can be written as a fraction \(\frac{
Rollout 7
0). Their new scores are original+5 (<=64). Average should be47. So

UP_PROJ

Top 16 Positive Activations
Rollout 1
)/2 = (3d2 / 2)/2 = 3d2/4.\n\n〈
Rollout 11
). Thus, p= (t1 +1/t1 ) + (t2 +1/t2
Rollout 11
= \overline{a} + \overline{b} + \overline{c}
Rollout 12
Compute 1 - (t2 - ν2)/(1 - ν2) )\n\n= (1 -
Rollout 11
. Let's write c as 1/(ab). Then q becomes a/b + b/(1/(
Rollout 11
ab). So\n\n\overline{a b} = \frac{1}{a b} →
Rollout 11
a*b ) =1/(a^2 z). Hence, c =1/(a^2 z
Rollout 11
hold if abc = 1/(a b c ), implying (abc)^2 = 1, so
Rollout 11
, let's write c = 1/(ab). Then:\n\nTerm1: c/b2 =
Rollout 1
||2 + ||x'||2 - d2)/2.\n\nBut perhaps another way is expressing this by
Rollout 11
*1/(r2 e^{i2θ}) = 1. So product is 1.
Rollout 11
2 b. Therefore, q = a^2 c + b^2 a + c^2 b
Rollout 9
So indeed, (0,10 -50/h) is between (0,5) and
Rollout 1
(||x||2 - ||x'||2)/2.\n\nBut if we assume that ⟨x -
Rollout 6
(log n + log p(n)) = O(log n) (since p(n) is poly(n),
Rollout 11
)\n\nGrouping terms:\n\n= [a +1/a ] + [ a z +1/(a z
Top 16 Negative Activations
Rollout 9
area common to triangle \(GEM\) and square \(AIME\) is 80 square units.
Rollout 13
12\n\]\n\n\[\n\boxed{187.12}\n\]<|im_end|>
Rollout 5
181.25\n\]\nThe greatest integer not exceeding 181.25
Rollout 8
7440\).\n\nThe probability is \(\frac{7440}{30
Rollout 9
\boxed{25}\n\n<|im_start|>answer\nSquare \(AIME\) has sides of length 10
Rollout 3
{m/s}\).\n\n**Final Answer:**\n\[\n\boxed{169} \,
Rollout 13
.24\n\]\n\n**Difference:**\n\[\n9628.24 -
Rollout 8
one block of red and one block of green.\n\n1. **Case 1: All red or all
Rollout 7
), \(\boxed{24}\), and \(\boxed{36}\).\n\n(b) No
Rollout 8
40}{30240} = \frac{31}{126}\).
Rollout 13
187.12\n\]\n\n\[\n\boxed{187.12
Rollout 6
{bad-angel}} \) unless \( \textbf{NP} = \textbf{P
Rollout 1
} y \text{ exists as described}}\n\]<|im_end|>
Rollout 12
0 < \nu < 1 \).\n\n\[\n\boxed{f(\nu) \text
Rollout 9
\) has sides of length 10 units. Isosceles triangle \(GEM\) has base
Rollout 8
\[\n\boxed{157}\n\]<|im_end|>

DOWN_PROJ

Top 16 Positive Activations
Rollout 7
5-point increase, every participant's score is increased by 5. Therefore, the new total score
Rollout 1
/ 2)/2 = 3d2/4.\n\n〈y, y〉 = ||y
Rollout 11
2θ} + e^{-iθ}\n\nRight side: 4 e^{-iθ} +
Rollout 8
from 1 to 4 reds and 4 to 1 greens).\n\nWait, so for each
Rollout 6
can query whether $s \in S$ and get the correct answer in return in constant time
Rollout 11
} + r3 ( e^{-iθ} - e^{iθ} ) + (1/(
Rollout 14
/d) - d*( (c/d)^2 /2 ) ) = c^2 /d -
Rollout 7
71P + 5P = 76P, and for repeaters, their total becomes
Rollout 4
by 5025:\n\nFirst, 5025 * 1200 =
Rollout 11
}\n\nc/a = (1/r2)/(r e^{iθ}) ) =1/(r3
Rollout 11
). Then q becomes a/b + b/(1/(ab)) + (1/(ab))/a.
Rollout 15
will be on.\n\nTherefore, next state is OOOTO... OTO? Wait, L1 on,
Rollout 1
.\n\nHence, from this, we can solve for ⟨x, x'⟩:\n\n⟨x,
Rollout 7
and the average of the non-promoted (repeaters) becomes 59.\n\nPart (a)
Rollout 4
:\n\nFirst, 5025 * 1200 = 6,030
Rollout 5
125\n\nFactor the second term:\n\nLeft side = (x + 175) -
Top 16 Negative Activations
Rollout 10
7≡2 mod5\n\nSo actually, the losing positions are numbers congruent to 0 or
Rollout 10
uent to 0 or 2 modulo5 are losing positions. Now, how to check if this pattern
Rollout 10
,1,2,3,...\n\nTherefore, the losing positions are numbers of the form 5k and
Rollout 10
of numbers n_i where each n_i is a losing position, and then show if we can construct n
Rollout 10
. Hmmm. If this continues, then the losing positions are numbers formed by 2,3,
Rollout 8
2 * [P(5, k) * P(5, 5 - k)] where k
Rollout 5
0)] / 2 = [150 ± sqrt(72500)] /
Rollout 14
q, r with 1/p +1/q +1/r =1. If we can split f
Rollout 10
roring.\n\nAlternative approach: The key is that each losing position forces the next player to enter a winning position
Rollout 8
example, if k=2, the number of ordered hands would be:\n\nFirst choose 2 red cards
Rollout 10
. Hmm. Let me list them again:\n\nLosing positions (P-positions): 0,
Rollout 8
10 different cards, hence the number of ordered sequences is 10P5. If they
Rollout 14
$ such that $1/p + 1/q + 1/r = 1$, then:\n\n$$
Rollout 14
$I^3 \leq 8 J K$.\n\nHmm. So, the left-hand side
Rollout 10
end up outside the set, and any number outside the set can be moved into the set.\n\nBut how
Rollout 10
position. Hence, n=17 is a losing position.\n\nBut previously, the pattern was alternated

Layer 39

GATE_PROJ

Top 16 Positive Activations
Rollout 3
calculate the mean molecular speed of radon gas. Hmm, first, I remember there's a formula for
Rollout 4
00 ± 0.02 V. Hmm, alright, let's start by recalling how to
Rollout 2
, is divided by 1000. Hmm, okay, let's unpack this step by step
Rollout 13
interest compounds quarterly compared to if it compounds annually. Hmm, I need to calculate the amount owed in both
Rollout 2
-3 \equiv 5 \mod 8\n \]\n\n2. **Modulo 1
Rollout 2
\]\n\n2. **Modulo 125 Calculation:**\n - Each term \(10
Rollout 11
}{c c c} $p=(a+b+c)+\left(\dfrac 1a+\
Rollout 2
\equiv 5 \mod 8\n \]\n\n2. **Modulo 125
Rollout 2
we can exploit when multiplying numbers modulo 1000.\n\nAlternatively, since 1000
Rollout 12
0 < \nu < 1 \). Hmm, let's try to figure out how to approach
Rollout 14
right)^3 \leq 8\left(\int_0^\infty f(x)^2
Rollout 4
_{1}=2000 \pm 10 \mathrm{pF}$ and $C
Rollout 2
^k - 1\), which modulo 8 is \(2^k - 1\).\n
Rollout 2
}_{\text{999 9's}}$ is divided by $1000$
Rollout 11
find all possible ordered pairs (p, q). Hmm, complex numbers with product 1. So abc
Rollout 9
triangle \(GEM\) and square \(AIME\) is 80 square units. We need to
Top 16 Negative Activations
Rollout 3
this question is ambiguous. Nevertheless, likely answer \boxed{170} m/s. Alternatively,
Rollout 3
the mean molecular speed of radon gas is \boxed{170} m/s if we round
Rollout 3
m/s. Thus, present answer boxed as \boxed{169 \text{ m/s}}
Rollout 3
50)=161.4, as 161^2=25,9
Rollout 3
K (unless stated otherwise), perhaps answer around \boxed{169} m/s. But need
Rollout 2
125 mod8: 125 /8= 15*8=12
Rollout 3
, the mean molecular speed of radon gas is \boxed{170} m/s if we
Rollout 9
=25. Thus, the answer is \boxed{25} . Seems done.\n\nThat's
Rollout 3
m/s.\n\nTherefore the valid boxed answer is \boxed{169}.\n\nBut initially I thought
Rollout 8
0. Let's see: 240×30 =7200, 7
Rollout 3
, this question is ambiguous. Nevertheless, likely answer \boxed{170} m/s. Alternatively
Rollout 7
0. Therefore, x can vary from 0 up to N/3. Since x= N/
Rollout 3
8,408). Let's see.\n\n168^2 = (170 -
Rollout 3
square root method:\n\n168.0^2=28,224.0\n\n
Rollout 3
). Let's see.\n\n168^2 = (170 - 2)^2 =
Rollout 3
0, which is incorrect. So correct is \boxed{169} m/s.\n\n**Final

UP_PROJ

Top 16 Positive Activations
Rollout 8
\(240 + 7200 = 7440\).\n\nThe probability is
Rollout 8
2400 + 1200 = 7200\).\n\nAdding both cases
Rollout 12
* [ sqrt( (t2 - ν2)(1 - t2) ) / (1 -
Rollout 4
{eq}}} = \frac{4.327}{1200} \approx
Rollout 8
+ n = 31 + 126 = 157\).\n\n\[\n\
Rollout 8
2400 + 1200 = 7200\).\n\nAdding both
Rollout 5
. The length \( x \) is given by:\n\[\nx = \sqrt{\frac{a
Rollout 12
can express \( f(\nu) \) as:\n\n\[\nf(\nu) = K(\sqrt
Rollout 4
.96 + 5.76} = \sqrt{18.72} \
Rollout 14
}.\n\]\n\nCombining these results, we get:\n\n\[\nI \leq \sqrt{J
Rollout 12
θ = sqrt( (t2 - ν2)(1 - t2) ) / (1 -
Rollout 4
.16 \cdot 15)^2} = \sqrt{3.6^2 +
Rollout 5
0 ± 10*sqrt(725)] / 2 = 75 ±
Rollout 4
Percentage error} = 0.008775 \times 100 \approx
Rollout 12
:\n\n= sqrt( (t2 - ν2)(1 - t2) ) / t\n\nThus
Rollout 4
16 \cdot 15)^2} = \sqrt{3.6^2 +
Top 16 Negative Activations
Rollout 8
, examples include RRRRR and GGGGG.\n\nSo actually, if all red cards in the layout
Rollout 9
0,0) lies within the square? Yes.\n\nWait, but then the area of the triangle in
Rollout 9
), where h is the altitude. Or is the altitude segment inside the triangle? Wait. Wait, in
Rollout 9
with base EM. Since triangle GEM is isosceles, meaning legs GE and GM are equal
Rollout 9
- h, y). But the triangle being isosceles with base EM. Since triangle GEM
Rollout 8
of the cards in a row in a random order."\n\nWait, the problem doesn't specify whether the cards
Rollout 9
the triangle being isosceles with base EM. Since triangle GEM is isosceles,
Rollout 0
fraction as part of between 0 and 1. When written in lowest terms, numerator a and denominator
Rollout 9
so the triangle is completely within the square. Wait, because the square spans from x=0 to x
Rollout 9
. Then the triangle GEM overlaps the square, and their common area is 80 square units.
Rollout 9
so overlap area is 50, as before. So how can we have overlapping area of 8
Rollout 9
triangle is within. But then how come the overlapping area is 80 then? Because that that case
Rollout 9
base of the triangle. Wait, no. Wait, EM is the top side of the square. Wait
Rollout 8
just says 5 red and 5 green cards.\n\nIs each red card identical and each green card identical
Rollout 9
the triangle? Wait. Wait, in an isosceles triangle, the altitude from the apex to
Rollout 11
but the problem states that p and q are real. Hence,p =s +t and q are real

DOWN_PROJ

Top 16 Positive Activations
Rollout 1
d / sqrt(2) times sqrt(2) from center y, which is exactly d. Wait
Rollout 15
0.\n\nThis is equivalent to x_{t+1}[i] = (x_{t}[
Rollout 15
can model it as a linear algebra problem over GF(2). Let me think.\n\nGiven that the next
Rollout 15
Else, if M is not nilpotent over GF(2), then there's a state that cycles
Rollout 11
^{-i2θ} - 4 e^{-i2θ}) + (e^{-iθ}
Rollout 0
b = 20!\) and \(a < b\).\n\nNow, how do I count such
Rollout 11
[ r^4 + r2 +1/r2 +1/r^4 ]\n\nTherefore, p=
Rollout 4
sqrt(7.696 x 10^-5). sqrt(7.696
Rollout 8
/30240=744/3024=744÷2
Rollout 9
area between these two lines is the area of trapezoid with bases at x=0 (length
Rollout 13
:\n\nActually:\n\n1.1255\n*1.1255\nMultiply as numbers
Rollout 13
11255\n----------\nMultiply 11255 by 11255
Rollout 15
, which is (x_{i-1} + x_{i+1}) mod2.\n\nTherefore
Rollout 1
all be distance d / sqrt(2) times sqrt(2) from center y, which is exactly
Rollout 10
10\n\nResult: L W L W W L W L W W L\n\nWait, okay
Rollout 5
and x with height h has area (75 + x)/2 * h = 62.
Top 16 Negative Activations
Rollout 3
/337≈0.047. Hence, 168.5 +0
Rollout 12
(ν2 + (1 - ν2)u2), but that might complicate.\n\nAlternatively,
Rollout 15
be the minimal k such that M^k = 0.\n\nHowever, I might not remember all details
Rollout 12
formula is -F(ν, ν) * 1, which is -∞, the actual result
Rollout 10
); mex{0}=1→Grundy[11}=1\n\nn=12: subtract
Rollout 4
00)/(10,000,000) = 0.003
Rollout 1
||x_i - y|| = d / sqrt(2).\n\nSo if such a y exists, then
Rollout 4
4*10^6)/(25*10^6)=4/25=0
Rollout 14
:\n\n\[\n\boxed{ \left(\int_0^\infty f(x) \, dx
Rollout 0
19 each have exponent 1 in 20!.\n\nSo exponents are: 18
Rollout 7
<40. 36 is less than 40, so all three are acceptable.\n\nWait
Rollout 10
subtract 1=14 (win), subtract 4=11 (win), subtract 9
Rollout 8
it's P(5,3) * P(5,2) for the green and red cards
Rollout 8
so sum m +n= 31 +126= 157.\n\n**Final
Rollout 14
\, dx \right)^3 \leq 8\left(\int_0^\infty
Rollout 5
Divide both sides by H (assuming H ≠ 0):\n\n(x + 175) -

Layer 40

GATE_PROJ

Top 16 Positive Activations
Rollout 13
351936.\n\nBreaking it down:\n\n6000*1 = 60
Rollout 13
36\n\nCalculating that value:\n\nFirst, 1 * 6000 = 6
Rollout 13
51936\n\nCalculating that value:\n\nFirst, 1 * 6000
Rollout 4
,150 by 5025:\n\nFirst, 5025 * 1
Rollout 8
divided by 240. Let's see: 240×30 =72
Rollout 8
0 divided by 240. Let's see: 240×30 =7
Rollout 13
12 multiplied by 1.12: \n\n1.12\nx1.12
Rollout 3
(28,436). Let's see, 168^2 = 2
Rollout 13
351936 = ?\n\nCompute separately:\n\n0.07351936
Rollout 13
1936\n\nCalculating that value:\n\nFirst, 1 * 6000 =
Rollout 4
327 divided by 1200.\n\nDivide by 100: 4
Rollout 13
50176\n\nAdding all these up:\n\n1.2544 + 0.
Rollout 13
936 * 6000.\n\nFirst, multiply 6000 by 1
Rollout 4
5025 �� let's divide 6,060,150 /
Rollout 13
\n\nSo multiply step by step, line by line:\n\n 1.1255\n
Rollout 4
/ 5025. \n\nDividing 6,060,150 by
Top 16 Negative Activations
Rollout 10
impartial game under the normal play condition is equivalent to a nimber. Also, all subtraction games have periodic
Rollout 1
a Hilbert space such that the images are orthonormal.\n\nBut here, we're dealing within the
Rollout 1
'⟩ = (2||x||2 - d2)/2.\n\nBut we don't know if
Rollout 12
ν, it's formally ( (ν2 - ν2) * (1 - ν2) )
Rollout 10
have periodic Grundy numbers if the set of allowed moves is finite. However, in this case, the
Rollout 10
every impartial game under the normal play condition is equivalent to a nimber. Also, all subtraction games have
Rollout 1
x||2 + ⟨y, y⟩ - d2 / 2.\n\nBut if ||x||
Rollout 12
a^2)(b^2 - x^2) ) ] dx\n\nWhich is a standard form
Rollout 12
the derivative \( \frac{dk}{d\nu} \) is:\n\n\[\n\frac{
Rollout 14
inequality called the Heisenberg uncertainty principle which relates position and momentum variances – not exactly the same,
Rollout 12
2 - ν2) becomes (ν2 - ν2) =0, so that first term would
Rollout 10
any two ni and nk, the difference is not a square. Such a set could be constructed using the
Rollout 1
S\right\}\] \nis an orthonormal system of vectors in $ \mathcal{
Rollout 1
bert space, one can have uncountable orthonormal systems.\n - The intersection of infinite closed
Rollout 12
( (t2 - ν2)(1 - t2) ) / (1 - ν2 )\n\n
Rollout 1
\n\[ \left\{\frac{\sqrt{2}}{d}(x\minus{}y):

UP_PROJ

Top 16 Positive Activations
Rollout 15
two.\n\nCheck with n=1 (power of two: 2^0), works.\n\nn=
Rollout 0
(even), 2 (7) (even?), no 7 has exponent 2? Wait for
Rollout 6
, not the other way.\n\nWait, but the problem doesn't say that k is given as part of
Rollout 15
, no: at t=1, L3 is off, L4 is on. Wait, no
Rollout 15
0, and others must not be included. But problem says "positive integers n ≥1”, and As
Rollout 11
, so that would mean one is real. But problem states none are real. Hence, that's impossible
Rollout 15
n=1 is 2^1 -1. Let's try n=3, which is
Rollout 15
the system doesn't die out. But we were supposed to find n's where regardless of the initial configuration
Rollout 11
that all variables are real? Wait, but the problem states that none of a, b, c are
Rollout 15
-1. Let's try n=3, which is 2^2 -1: no,
Rollout 15
.\n\nWait, but L2 in t=2 is on, L1 is off, L3 is
Rollout 8
* 63=2*7*9. 31 and 126 are cop
Rollout 15
2 minus 1? Wait, n=1 is 2^1 -1. Let's try
Rollout 9
,0). Wait, wait no: in the problem statement, the square is called AIME. If
Rollout 15
be a power of2? BUT n=1 is OK (2^0), and n=2
Rollout 15
L2 (off). So it turns off. L2: neighbors L1 (on) and L
Top 16 Negative Activations
Rollout 1
y. So the main idea is to find a point y such that all x in S are at distance
Rollout 1
2).\n\nSo, how do we show that such a y exists?\n\nIn finite dimensions, if I have
Rollout 4
about the percentage error in the energy, I need to figure out how the uncertainties in C1, C
Rollout 4
the energy stored in a capacitor is (1/2) * C_eq * V2. Since the
Rollout 5
, if there is a segment parallel to the bases that divides the trapezoid into two regions of
Rollout 13
by step.\n\nFirst, let's recall the formula for compound interest. The formula is A = P(
Rollout 11
of them are real, and none have absolute value 1. They define p as (a + b
Rollout 11
). \n\nLet me think of a different approach. Let me let a, b, c be three numbers
Rollout 1
d.\n\nNow, the problem wants me to find a point y such that when I take (sqrt(
Rollout 13
think step by step.\n\nFirst, let's recall the formula for compound interest. The formula is A =
Rollout 5
the lengths involved. The formula for the length of a line segment parallel to the bases that creates a certain
Rollout 1
orthonormal.\n\nSo the key is to find a point y such that all x in S lie on
Rollout 1
||x - y||2 = d2 / 2.\n\nExpanding, that gives:\n\n⟨x
Rollout 1
Therefore, ⟨x - y, x' - y⟩ = 0.\n\nSo another way to think
Rollout 1
thonormal.\n\nSo the key is to find a point y such that all x in S lie on a
Rollout 4
given by \( E = \frac{1}{2} C_{\text{eq}} V^

DOWN_PROJ

Top 16 Positive Activations
Rollout 1
1 + r2 >= l and |r1 - r2| <= l.\n\nIn our case,
Rollout 14
up a functional:\n\n$$\n\mathcal{L} = \left( \int f(x)
Rollout 14
\int f(x) dx \right)^3 - \lambda \left( \int f(x)^
Rollout 14
functional to maximize is ��f dx -λ ��f2 dx - μ ��x f
Rollout 14
, we can set up a functional:\n\n$$\n\mathcal{L} = \left( \
Rollout 14
�f dx -λ ��f2 dx - μ ��x f dx.\n\nTaking the functional
Rollout 14
\int f(x)^2 dx - C \right) - \mu \left( \int x
Rollout 11
+ b + c) + (b + c) + bc. Wait, not sure. Alternatively,
Rollout 9
5)/h.\n\nThe equation is y -0 = (-5)/h (x -10).
Rollout 1
if we have a family of closed convex sets where every finite intersection is non-empty, then the whole intersection
Rollout 14
int f(x) dx \right)^3 - \lambda \left( \int f(x)^2
Rollout 0
2) + floor(20/4) + floor(20/8) + floor(
Rollout 14
f(x)^2 dx - C \right) - \mu \left( \int x f(x
Rollout 11
ab + bc + ca). So that's correct. So p = (a + b + c)
Rollout 8
can be simplified. Since 31 is prime. 126 divides by 2,
Rollout 0
product of two coprime numbers, \(a\) and \(b\), with \(a < b
Top 16 Negative Activations
Rollout 14
||f(x)x^(-1/3)||_a *||x^(1/3)||_
Rollout 14
\cdot 1 \le \|f\|_a \|1\|_b,\n\nBut as
Rollout 4
V separately.\n\nStarting with C1: 2000 ± 10 pF. The
Rollout 14
Then Hölder gives:\n\nI ≤ ||f||_2 ||1||_4 ||1||_
Rollout 14
||f(x)x^{1/2} ||_2 * ||x^{-1/2}||
Rollout 14
^{-1/3}| dx ≤ ||f||_3 ||x^{1/3}||_
Rollout 4
025. \n\nDividing 6,060,150 by 50
Rollout 14
_a *||x^(1/3)||_b,\n\nwith 1/a +1/b=
Rollout 11
of a different approach. Let me let a, b, c be three numbers such that abc =
Rollout 14
I ≤ ||f||_2 ||1||_4 ||1||_4.\n\nBut ||1
Rollout 11
similarly if r <1, r3 +1/r3 >2. Wait, Wait, let's
Rollout 4
25 �� let's divide 6,060,150 / 50
Rollout 14
2 * ||x^{-1/2}||_4 * ||1||_4.\n\nHmm:\n\n
Rollout 14
12^{2/3}=(12^{1/3})2=( approximately 2.
Rollout 5
100$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. I need to solve
Rollout 14
2K)/ J^{1/4}= J^{1/2}*(2K)^{1

Layer 41

GATE_PROJ

Top 16 Positive Activations
Rollout 3
8 K (25°C) yields ~168.5 m/s. Therefore, since the
Rollout 3
and rounded to three sig figs is 169 m/s. Thus, present answer boxed as
Rollout 3
compute", so it's expected to compute.\n\nAlternatively, perhaps check if there is an established average speed for
Rollout 3
precision needed.\n\nAlternatively, maybe the original question assumes a specific temperature but forgot to mention. If the user
Rollout 3
and calculator computations, perhaps the answer expects 170 m/s as an approximate answer. But some
Rollout 3
98 K (25°C) yields ~168.5 m/s. Therefore, since
Rollout 3
. Since the user hasn't specified temperature, perhaps there's a standard temperature to assume. Let me check
Rollout 3
different.\n\nBut given that the user hasn't specified the temperature, this seems ambiguous. However, in many
Rollout 11
them as a, b, c such that one is the product of the other two. But maybe no
Rollout 3
is rounded to three significant figures, giving \(169 \, \text{m/s}\).\n\n
Rollout 3
around 298 K. Maybe. Alternatively, as room temperature sometimes is taken as 30
Rollout 3
's expected to compute.\n\nAlternatively, perhaps check if there is an established average speed for radon gas at
Rollout 3
, given that no temperature is given, perhaps they just want the formula, but I don't think so
Rollout 3
, if forced to choose, answer approximately 170 m/s. But given compute precisely, it
Rollout 11
=1, perhaps consider that one of the variables is the product of conjugates of the others. For
Rollout 3
m/s. But if the question hasn't specified the temperature, this can vary.\n\nAlternatively, given that
Top 16 Negative Activations
Rollout 13
3000 → 0.57 * 6000 = ?\n\nWait, maybe
Rollout 13
0.57 * 6000 = ?\n\nWait, maybe better to process 60
Rollout 3
Let's see.\n\n168^2 = (170 - 2)^2 =
Rollout 13
2: 1*1 = 1, 1*0.12 = 0.
Rollout 8
0×30 =7200, 7440-7200=
Rollout 13
2.\n\nWait, let me check that subtraction:\n\n9,628.24\n\n-9
Rollout 13
squared. Let's break it down:\n\nFirst, 1 * 1.2544 =
Rollout 13
x1.12\n--------\n224 (1.12 * 2)\n11
Rollout 4
,030,000, since 5025*1000=
Rollout 4
24=14,328 and 597*0.8004
Rollout 13
me check that subtraction:\n\n9,628.24\n\n-9,441.
Rollout 13
000 = 6000\n\n0.5 * 6000 =
Rollout 13
11616 = ?\n\nCompute that:\n\n9628.23864 -
Rollout 13
,441.12\n\n------------\n\nSubtract dollars: 9628 -
Rollout 8
0. Let's see: 240×30 =7200, 7
Rollout 8
by 240. Let's see: 240×30 =720

UP_PROJ

Top 16 Positive Activations
Rollout 0
but their exponents are 1, which is odd. Actually, wait nvm, the exponents
Rollout 9
). The altitude is h, the horizontal distance from G to EM. So h is the length we need
Rollout 2
7 terms) is-1 mod1000. So multiplying 891 by (-1
Rollout 10
win. Wait, 2 is a losing position, so if first player moves from 3 by taking
Rollout 0
:4,7:2, which are all even? Wait, 18,8,4
Rollout 4
, so 0.88% is already two significant figures. Alternatively, if written as 0
Rollout 1
= sqrt(2) * ||x - y|| / sqrt{1 - a^, but this
Rollout 9
care. So GEM is an isosceles triangle with base EM, so EM is from (
Rollout 15
, next state (t=1): O T O.\n\nNow, check each lamp:\n\n- L1
Rollout 0
128\), since each pair is counted twice. But actually, since we are considering ordered pairs
Rollout 15
will turn off. If it's off, it stays off. Therefore, regardless of the starting state (
Rollout 15
sum 0 or 2, it's 0. Therefore, XOR gives the same result as modulo
Rollout 0
exponent of 7 is 2, which is even. Wait a second. Then primes 11
Rollout 1
- y) : x ∈ S } are orthonormal. Hence, they can be considered as an
Rollout 1
the vectors from the centroid to the vertices are orthonormal. Wait, not exactly, but in the
Rollout 0
each unordered pair {a, b} is counted twice in the 256 ordered pairs.\n\nBut
Top 16 Negative Activations
Rollout 10
infinitely many positions exist. As a contest problem, the solution may be found in realizing the periodicity failed
Rollout 6
time.\n\nSo perhaps the answer is not straightforward, hence why reference to the P/poly theorem. The
Rollout 14
Either way, this isn't helping. Perhaps I need a different approach.\n\nAlternative idea: Use Hölder
Rollout 15
the possible patterns and their evolutions. Perhaps a solution exists where nが is a power of two?
Rollout 15
, power configurations may make difference.\n\nAlternatively, perhaps answer is that n must be a power of2
Rollout 14
the person who set the problem must have an particular method in mind. Let's go back.\n\nAnother strategy
Rollout 6
in circles here. Given the time constraints, maybe for problem 3, since it's known in complexity
Rollout 14
who set the problem must have an particular method in mind. Let's go back.\n\nAnother strategy. Use
Rollout 10
exist. As a contest problem, the solution may be found in realizing the periodicity failed, but that
Rollout 1
only countable ones.\n\nBut the question is probably supposed to allow the System S to be mapped to or
Rollout 6
P/poly.\n\nTherefore, for the first problem, combining multiple sparse sets into one via tagging. For
Rollout 10
positions exist. As a contest problem, the solution may be found in realizing the periodicity failed, but
Rollout 14
person who set the problem must have an particular method in mind. Let's go back.\n\nAnother strategy.
Rollout 1
the subject.\n\nAlternatively, but, finally, the answer is asserting that such a y exists because he’s
Rollout 10
. As a contest problem, the solution may be found in realizing the periodicity failed, but that a
Rollout 6
problem asks to show it, not just cite, perhaps construction:\n\nGiven L ∈ P_angel, then

DOWN_PROJ

Top 16 Positive Activations
Rollout 11
1, 1/a = bc, 1/b = ac, 1/c = ab, as
Rollout 11
Hmm. So q = a^2 c + b^2 a + c^2 b. Since
Rollout 4
sqrt(0.000064) is 0.008, and sqrt
Rollout 11
. Therefore, q = a^2 c + b^2 a + c^2 b. Hmm
Rollout 4
=30,150. Then, 30,150 /502
Rollout 11
can write 1/a = bc, 1/b = ac, 1/c = ab. So
Rollout 11
= ab. So 1/a + 1/b + 1/c = bc + ac + ab
Rollout 11
. But then abc = (1/b)(1/c)(1/a) = 1/(a b
Rollout 11
is (a + 1/a) + (b + 1/b) + (c +
Rollout 4
0, since 5025*1000=5,025,0
Rollout 13
0.\n\nFirst, multiply 6000 by 1.5735193
Rollout 11
is not 1. So all of a, b, c are complex but non-real, with modulus
Rollout 11
Then indeed, if a = 1/b, b = 1/c, c = 1/a
Rollout 4
14,328 and 597*0.8004≈ 5
Rollout 13
0.57351936\n\nWait, or perhaps it's easier to multiply
Rollout 11
1=0, but they are 1 and the complex roots. But those have modulus 1.
Top 16 Negative Activations
Rollout 4
1200. Yep, that's right. So, C_eq is 120
Rollout 10
between 3 and 2? Let's see:\n\n0 to2: +2\n\n2 to5
Rollout 9
,0)):\n\nThe slope (m) is (0 -5)/(10 - (10
Rollout 14
{w(x)} f(x) dx \leq \left( \int_0^\infty
Rollout 3
8≈168, which is about right. So, the number looks consistent. Therefore, if
Rollout 4
respectively. Then, error terms calculated for each. \n\nAlternatively, perhaps I could use logarithmic differentiation.
Rollout 2
..., up to k=999 (10^999 -1). Therefore, the
Rollout 8
,2) = same 60*20= 1200. Times 2
Rollout 10
Let's see:\n\n0 to2: +2\n\n2 to5:+3\n\n5 to7:+
Rollout 10
1}=2→Grundy[19}=2\n\nn=20: subtract1→1
Rollout 9
0 - (10 - h)) = (-5)/h.\n\nThe equation is y -0 =
Rollout 8
5*4)=2*60*20=2400\n\nk=4:\n\n
Rollout 8
sequences: R GGGG, GGGG R\n\nk = 2: RR GGG,
Rollout 14
λ(x) = λ^{a} f(λ x). Then:\n\nCompute I_λ = �
Rollout 10
moves to W. So n=20:L.\n\nn=21: n -1=2
Rollout 4
, etc., would give the same result as before.\n\nLet me actually compute it this way:\n\nFirst term

Layer 42

GATE_PROJ

Top 16 Positive Activations
Rollout 13
4 years. How much more money will she owe if the interest compounds quarterly than if the interest compounds
Rollout 6
{N} \mapsto \mathbb{N}$ such that for each $n \
Rollout 6
S$ is bounded by $p(n)$ . In other words, $|S^{
Rollout 12
\] \n(where the positive value of the square root is taken) is monotonically decreasing in the interval
Rollout 1
orthogonal.\n\nFirst, let me recall that an orthonormal system requires two things: 1) each
Rollout 10
every impartial game under the normal play condition is equivalent to a nimber. Also, all subtraction games have
Rollout 14
0,\infty)\to[0,\infty)$ is integrable and $$ \int
Rollout 15
$n\geq1$ be a positive integer. $n$ lamps are placed
Rollout 15
$ can we guarantee that all lamps will be off after some time?\n<|im_start|>assistant\n<|im_start|>think\n
Rollout 6
} \mapsto \mathbb{N}$ , a sequence of strings $\{\alpha_n\
Rollout 6
N} \mapsto \mathbb{N}$ such that for each $n \in
Rollout 6
|S^{=n}| \leq p(n)$ , where $S^{=n}
Rollout 1
S\right\}\] \nis an orthonormal system of vectors in $ \mathcal{
Rollout 8
$m$ and $n$ are relatively prime positive integers. Find $m + n$ .\n<|im_start|>
Rollout 5
into two regions with areas in the ratio 2:3. Then, we need to find x,
Rollout 14
p, q>1 with1/p +1/q=1,\n\nab≤ (a^p)/
Top 16 Negative Activations
Rollout 8
), or all reds or all greens.\n\nSo first, we can calculate the number of such happy sequences
Rollout 8
0P5).\n\nLet me try this approach.\n\nFirst, how many color sequences are "happy"?\n\nCase
Rollout 4
9 J=15 nJ.\n\nSimilarly, compute E_max=0.5*120
Rollout 13
351936.\n\nBreaking it down:\n\n6000*1 = 60
Rollout 8
P(5,5 -k)]\n\nLet me compute this for k=1 to k=4.\n\n
Rollout 13
this:\n\nFirst, 6000 * 1.6 = 9600\n\n
Rollout 8
30240.\n\nNow, we need to count how many of these sequences have all red cards
Rollout 11
re^{iθ})\n\nLet me compute step by step:\n\n- a + b = 2r e
Rollout 13
7351936.\n\nBreaking it down:\n\n6000*1 = 6
Rollout 13
3736\n\nLet's compute this:\n\nFirst, 6000 * 1.
Rollout 9
). Connecting (10,0) to (10,10), to (0,1
Rollout 8
First, how many color sequences are "happy"?\n\nCase 1: All reds (RRRRR
Rollout 3
22 kg/mol.\n\nNow, the average speed is sqrt(8RT/(πM)).\n\nBut
Rollout 13
6 step by step:\n\n6000 * 1 = 6000\n\n60
Rollout 3
))\n\nLet's calculate the numerator and denominator step by step.\n\n8*R*T = 8 * 8
Rollout 13
3736:\n\n6000 * 1 = 6000\n\n60

UP_PROJ

Top 16 Positive Activations
Rollout 8
31/126, so m=31, n=126.\n\nWait
Rollout 5
the greatest integer not exceeding that is 181. Therefore, the answer is 181
Rollout 8
126, so m=31, n=126.\n\nWait, but let's
Rollout 5
181.25 is 181. Therefore, final answer is 181
Rollout 5
the greatest integer not exceeding this is 181.\n\nWait, this seems straightforward now, but maybe
Rollout 8
26, so m=31, n=126.\n\nWait, but let's confirm
Rollout 13
nearest cent, is $187.12\n\nSo, all right, confirms that the difference
Rollout 9
=25. Thus, the answer is \boxed{25} . Seems done.\n\nThat's
Rollout 5
1. Therefore, the answer is 181. \n\nWait, but let me verify once again
Rollout 5
is 181. Therefore, the answer is 181. \n\nWait, but let
Rollout 4
ounding to two decimal places, 0.88%. However, the question says "percentage error",
Rollout 1
ram law says that ||u + v||2 + ||u - v||2 = 2||
Rollout 4
74%, which rounds to 0.88%. Wait, let me check the exact value.\n\n
Rollout 5
is 181. Therefore, final answer is 181. \n\nI think that's
Rollout 4
, the percentage error is approximately 0.88%, if we round to two decimal places. But
Rollout 3
this question is ambiguous. Nevertheless, likely answer \boxed{170} m/s. Alternatively,
Top 16 Negative Activations
Rollout 11
roots.\n\n3. **Symmetric and Real Conditions**:\n - Since \(p\) and \(q
Rollout 11
and cyclic in \(a, b, c\).\n\n2. **Transformation and Simplification**:\n
Rollout 14
Another idea: For each x, use the AM-GM inequality.\n\nSuppose for each x, we
Rollout 11
\) and analyze the symmetry and properties of these roots.\n\n3. **Symmetric and Real Conditions**:\n
Rollout 13
thousandths place shifted by four decimal places for multiplier position):\n0.056275\n
Rollout 14
in that manner.\n\nAnother approach. Think variables, exponents. We need to relate three things: L
Rollout 11
of two variables \(a\) and \(b\).\n - After several attempts to simplify and test specific
Rollout 1
able sequence, but assuming that S is any set. Hmm.\n\nAlternatively, since you can extend any or
Rollout 11
expressions must satisfy certain symmetry and conjugate properties. \n - After extensive analysis and considering the problem's
Rollout 14
geq A f(x) \), so:\n\n\[\nI_2 \leq \frac{
Rollout 10
several positions, showing non-periodic but structured behavior.\n\n4. **Constructing Infinite Losing Positions**:\n
Rollout 14
, the constant isn't optimal but achievable via AM-GM. Wait, the person who set the problem
Rollout 11
c = \frac{1}{ab}\), we express \(p\) and \(q\) in terms
Rollout 13
7 * 6000 = ?\n\nWait, maybe better to process 6000 *
Rollout 11
- After several attempts to simplify and test specific cases, we consider the possibility of the cubic polynomial whose
Rollout 11
\) and \(b\).\n - After several attempts to simplify and test specific cases, we consider the

DOWN_PROJ

Top 16 Positive Activations
Rollout 0
of such coprime pairs is \(2^k\), where \(k\) is the number of
Rollout 15
matrix is nilpotent if some power of it is the zero matrix.\n\nSo, for M being the
Rollout 4
Then, take derivatives:\n\ndC_eq / C_eq = (dC1 / C1 + d
Rollout 2
coprime. Then, using the Chinese Remainder Theorem, combine the results. But before jumping
Rollout 14
Wait, let's see. Let's denote:\n\n$I = \int_0^\infty f(x
Rollout 11
i sin3θ )\n\nSimplify real and imaginary parts:\n\nReal: 1 +8cos3θ
Rollout 14
in the following manner:\n\n$$ \left( \int f(x) dx \right)^3 = \
Rollout 2
9 again. 891*999 again: as before, 891*
Rollout 7
is twice the number of repeaters.\n\nTherefore, N = P + R = 2R + R
Rollout 14
mathcal{L} = \left( \int f(x) dx \right)^3 - \
Rollout 2
is5 mod8. So according to Chinese Remainder Theorem, there is a unique solution mod1
Rollout 12
. But not sure. Let me recall the Beta function:\n\nB(p, q) = ��_
Rollout 7
Since all participants are either promoted or repeaters, N = P + R.\n\nWe are given:\n\n1
Rollout 11
\n\nUsing Euler’s formula, e^{iΦ} - e^{-iΦ} = 2i
Rollout 9
,0) to (10, 10). Then EM is the side from E (1
Rollout 8
Wait, for R^k G^(5 -k):\n\nNumber of ways to choose and arrange k red
Top 16 Negative Activations
Rollout 14
$. So $f(x) = a - b x$. \n\nBut wait, we are assuming that $
Rollout 13
55\nMultiply as numbers:\n\n 11255 (ignore decimals for now)\nx
Rollout 14
8*(1/(2k))*(1/k^2 ))= 1/(8*(1/(2
Rollout 5
12500 = 125y\n\nSimplify:\n\ny^2 - 2
Rollout 4
060,150 / 5025 �� let's divide 6,
Rollout 14
dx ) * sqrt(∫A^infty 1 dx )= sqrt(J2)* sqrt(in
Rollout 14
als:\n\n∫(f(λ x))2 λ dx scales as μ2 λ ��f2
Rollout 4
.000077 is 7.7e-5. sqrt(7.7
Rollout 10
: losing.\n\nn=6: possible moves: 1, leaving 5 (losing). So n
Rollout 2
10^1 - 1\n- 99 = 10^2 -
Rollout 6
given |x|, which is the length of x, then the combined machine would first compute α_{
Rollout 15
],\n\n [0,1,0],\n\n [1,0,1]]\n\nM^3 =
Rollout 10
to a losing position).\n\nn=4: from 4, you can move to 4-1
Rollout 5
- 12500 = 125y\n\nSimplify:\n\ny^2 -
Rollout 2
1)...(10^{999} -1). We need to compute P mod
Rollout 11
.= 1/z + (a z )*a2 z +1/(a3 z )\n\n=1

Layer 43

GATE_PROJ

Top 16 Positive Activations
Rollout 9
Final Answer**\n\boxed{25}\n\n<|im_start|>answer\nSquare \(AIME\) has sides of length
Rollout 0
Answer**\n\boxed{128}\n\n<|im_start|>answer\nGiven a rational number between 0 and
Rollout 8
Answer**\n\boxed{157}\n\n<|im_start|>answer\nKathy has 5 red cards and
Rollout 3
boxed{169} m/s.\n\n<|im_start|>answer\nTo compute the mean molecular speed \( v \
Rollout 5
Answer**\n\boxed{181}\n\n<|im_start|>answer\nLet the shorter base be \( b \)
Rollout 14
(x) \, dx \right) }\n\n<|im_start|>answer\nTo prove the inequality \n\n\[\n\left
Rollout 15
) for non-negative integers \( k \).\n\n<|im_start|>answer\nTo determine for which values of \( n \
Rollout 11
\boxed{(-1, 3)}\n\n<|im_start|>answer\nGiven three complex numbers \(a, b,
Rollout 13
boxed{187.12}\n\n<|im_start|>answer\nTo determine how much more money Joanie will
Rollout 12
on } (0, 1)}.\n\n<|im_start|>answer\nTo prove that the function \( f(\nu
Rollout 4
{0.88\%}.\n\n<|im_start|>answer\nTo find the percentage error in the calculation of
Rollout 10
the second player can force a win.}\n\n<|im_start|>answer\nTo prove that there are infinitely many initial situations
Rollout 7
36}\n\n(b) No solution exists.\n\n<|im_start|>answer\nGiven a mathematics test with \( N \)
Rollout 2
Answer**\n\boxed{109}\n\n<|im_start|>answer\nTo find the remainder when \(9 \times
Rollout 1
} y \text{ exists as described}}\n\n<|im_start|>answer\nGiven an infinite-dimensional Hilbert space \( \
Rollout 13
quarterly compounding (n = 4):**\n\[\nA_{\text{quarterly}} =
Top 16 Negative Activations
Rollout 11
a*b*c =k*k*(1/k2) = 1. Also, none of them are
Rollout 8
by dividing numerator and denominator by 240: 7440÷240=
Rollout 9
0] ( y_upper - y_lower ) dx\n\n= ��[0 to10] [
Rollout 11
invalid. Alternatively, maybe non-real c as well.\n\nAlternatively, take a = re^{iθ},
Rollout 3
8^2 = 28,224. 169^2 = 2
Rollout 11
= a^2 * (a b)/b = a^3\n\nTerm3: b2/(
Rollout 15
. So L3's next state is on.\n\nTherefore, at t=2, lamps L1 is
Rollout 8
, they each need to form a single block.\n\nSo if in the laid out sequence, there is at
Rollout 8
40÷240=126. As above. Therefore 31/12
Rollout 2
0^1 =10 mod125\n\n10^2=100 mod1
Rollout 8
0=1200\n\nk=2:\n\n2*P(5,2)*P(
Rollout 7
294/3=98 per student, but it's impossible since max original repeater is
Rollout 10
subtract 1: 11 (winning); 12-4=8 (winning
Rollout 15
So L2 is on at t=2. Therefore, t=2 is F O O.\n\nThen
Rollout 0
's either assigned entirely to a or to b.\n\nTherefore, 20! can be considered as a
Rollout 13
50176\n\nAdding all these up:\n\n1.2544 + 0.

UP_PROJ

Top 16 Positive Activations
Rollout 8
! ) / (5 - (5 -r))! ; wait, maybe it's better to think
Rollout 1
d / sqrt(2) times sqrt(2) from center y, which is exactly d. Wait
Rollout 13
ab + b^2 where a = 1, b = 0.1255\n\n
Rollout 15
values mod 2 being 1. But wait, if both neighbors are on, sum is 2
Rollout 1
of all these spheres (each centered at x ∈ S with radius d / sqrt(2)) is non
Rollout 1
simplex, in 3D, i.e., tetrahedron, but even a triangle is
Rollout 4
the 8 in the tenths place). Wait, percentage is per hundred, so 0.8
Rollout 1
0 for i ≠ j. But actually no, in the previous, if we take multiple points,
Rollout 8
5 * (5 *4 *3 *2 *1), but wait no:\n\nWait, for R
Rollout 2
^k-1 for anyk is always even? No. When k=1: 9 is
Rollout 9
Wait, EM is the top side of the square. Wait, hold on. \n\nWait, in the
Rollout 12
evaluated at t = ν] * dν/dν + ��_{ν}^1 ��
Rollout 0
b have exponents for the primes? Wait no, in the problem statement, the numerator and denominator (
Rollout 8
all 5 red cards in order), but wait, she is laying out 5 cards out of
Rollout 4
%, which is half of 1.76% (max-min). But clearly, max-min range
Rollout 1
the intersection of all these spheres (each centered at x ∈ S with radius d / sqrt(2))
Top 16 Negative Activations
Rollout 12
+ u)^2 - ν2 ) (1 - (ν + u)^2 ) )\n\nSimplify
Rollout 7
5*(2N/3) = (75*2)/3 *N=150
Rollout 4
25.2004=603*25 +603*0.
Rollout 7
)=79*(16 +2)=79*18=1422.\n\nNon
Rollout 9
] dx\n\n= ��[0 to10] [ (1/5 x +8 +
Rollout 13
:\n\nA_annual = 6000*(1 + 0.12)^4\n
Rollout 7
/3.\n\nSubstitute P and R:\n\n75*( (2N/3) + x )
Rollout 11
implify:\n\n= 4 e^{iθ} + 4 e^{i2θ} + (
Rollout 5
)/(75 + x)\n\nSo the equation becomes:\n\n(x + 175)(x -
Rollout 7
. Let's compute that.\n\nFirst expand:\n\n75*(2N/3) +75x
Rollout 7
denominator:\n\n62N/12 +54N/4 =62N/12
Rollout 7
- x.\n\nWe are told that after the increase:\n\n- The average of promoted becomes 75.
Rollout 7
=71N\n\nCalculate the terms:\n\n75*(2N/3) = (75
Rollout 3
73 �� 66.512 * 270 = 17,
Rollout 11
*(2i sinθ) + (15/4)*(2i sin2θ) =
Rollout 11
of ��_a, ��_b, ��_c is 0 or π. Hmm. Also

DOWN_PROJ

Top 16 Positive Activations
Rollout 5
5H\n2. (x + 175)(H - h) = 12
Rollout 14
.\n\nSo suppose x ranges up to x_max=1/μ, and for x ≤ x_max,
Rollout 3
) in m/s\n<|im_start|>assistant\n<|im_start|>think\nOkay, I need to calculate the mean molecular speed
Rollout 3
in m/s\n<|im_start|>assistant\n<|im_start|>think\nOkay, I need to calculate the mean molecular speed of
Rollout 9
10 is 100 -500/h.\n\nBut this is the overlapping area. Because
Rollout 5
5H\n2. (x + 175)(H - h) = 12
Rollout 9
, 10), M is at (10, 10), and E is at (
Rollout 5
-hand side:\n\ny^2 - 125y + 100y - 1
Rollout 9
] [5/h x +10 -50/h +5/h x -50/h]
Rollout 9
for an isosceles triangle with vertical base EM. Therefore, G is at (10 -
Rollout 3
16\n\nNow, 7,432.716 / 0.22
Rollout 4
combination of capacitors?\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to find the percentage error
Rollout 9
-1000/h\n\n= 500/h +100 -100
Rollout 9
A is the bottom-left corner, I is the top-left, M is the top-right, and E
Rollout 15
2: neighbors L1 (off) and L3 (on). Exactly one neighbor on. Therefore,
Rollout 9
2/2) + (10 -100/h)*x ] from 0 to1
Top 16 Negative Activations
Rollout 14
dx ) * sqrt(∫A^infty 1 dx )= sqrt(J2)* sqrt(in
Rollout 7
they combine with x students whose original scores were between 60-64 (so with original average
Rollout 6
� P. Conversely, P �� P_bad-angel because we can ignore the angel string (set
Rollout 2
10^3=1000 mod125. 1000 divided by
Rollout 9
0 -100/h)*x ] from 0 to10.\n\nEvaluating at
Rollout 5
) = 125H\n\nFrom equation 1: h = (125H)/(
Rollout 13
$9,441.12 after 4 years.\n\nNow let's compute the amount owed
Rollout 7
problem states that the pass mark was "fixed at 65". When all scores are increased by
Rollout 9
10 - y)^2\n\nCanceling h^2:\n\ny^2 = (10 -
Rollout 9
10? Wait, if h is less than10, then G is at (10 -
Rollout 3
5 +15.87/(2*168.5)=15.87
Rollout 3
314*298/(π*0.028)) �� sqrt( (
Rollout 2
00.\n\nHence same asmodulo8 and125approach.\n\nTherefore, two separate methods
Rollout 11
b, c are real, and none have modulus 1. So none of r_a, r_b
Rollout 3
8,157.776 / 0.697 �� 26
Rollout 11
| are all greater than 1 or less than 1. But problem states that none of them have

Layer 44

GATE_PROJ

Top 16 Positive Activations
Rollout 2
we consider the product modulo 8 and modulo 125 separately, then combine the results using the
Rollout 8
adjacent and all green cards laid out are adjacent.\n\nFirst, we calculate the total number of possible sequences of
Rollout 11
where none of them are real or have absolute value 1, we need to find the possible values of
Rollout 9
0)\), and \(E(10, 0)\).\n\n2. ** Isosceles
Rollout 11
\(p\) and \(q\) are real.\n\n1. **Key Observations**:\n - Since
Rollout 0
factors of 20! are 2, 3, 5, 7, 1
Rollout 2
99 9's}}\) is divided by 1000, we consider the product modulo
Rollout 3
radon in kg/mol.\n\n**Steps:**\n\n1. **Molar Mass of Radon (R
Rollout 0
a \times b = 20!\).\n\n1. **Prime Factorization of 20!
Rollout 9
EM\) in \(\triangle GEM\).\n\n1. **Vertices of the Square**:\n -
Rollout 15
answer\nTo determine for which values of \( n \) all lamps will eventually be off regardless of the
Rollout 3
\times 8.314 \times 298 = 19,82
Rollout 14
the integral \( I = \int_0^\infty f(x) \, dx \), \(
Rollout 3
\) of radon (Rn) gas, we use the formula for the mean (average)
Rollout 15
ly turns off.\n - For \( n = 2 \) and higher, testing specific configurations shows
Rollout 9
0, 0)\) to \(M(10, 10)\).\n - The
Top 16 Negative Activations
Rollout 8
/ (5 - (5 -r))! ; wait, maybe it's better to think:\n\nWait
Rollout 14
)^{1/2 } )^{.1 } + sqrt(J)/2.\n\n Wait, actual calculation
Rollout 1
squared / 2 - ) ) / 2.\n\nBut not making progress.\n\nWait but if subtract the
Rollout 11
2 = (1 + p) -2 =\n\nWait earlier p= (1 +a)(1
Rollout 0
, then 5, 2, 1.\n\nWait, 18 for exponent of 2
Rollout 12
= 1 - k2 cos2θ\n\nWait, hold on:\n\nWait, let's write nu
Rollout 10
2 modulo 2 and 5 modulo 5? Not sure. Alternatively, these numbers could be written
Rollout 12
- ν)^{-1/2} integrable.\n\nBut when we evaluate F at t=ν,
Rollout 10
7,20,22,25?\n\nWait, but n=25 is a losing
Rollout 9
50/h), G's projection path, etc. Wait, perhaps getting complicated.\n\nAlternatively, maybe split
Rollout 4
+ 0.000064 = 0.0000769
Rollout 4
% and 0.004% error, but the voltage uncertainty was ±0.02
Rollout 7
answer for part(b) is N=24?\n\nWait, wait. Earlier we saw that in equations
Rollout 12
2)(1 - t2) ) ] dt, but we have to check:\n\nWait, from the
Rollout 10
-0=0;0 -1=3;\n\nIf n ��1 mod4: subtract1
Rollout 6
poly(2^{m}) = exponential in m. Therefore, the set S_L contains exponentially many strings

UP_PROJ

Top 16 Positive Activations
Rollout 3
) in m/s\n<|im_start|>assistant\n<|im_start|>think\nOkay, I need to calculate the mean molecular speed
Rollout 8
7200=7440. Therefore, 7440 /10*
Rollout 15
O F\n\nOverlap, cycles every two steps. So discrepancies exist between GF(2) model and original
Rollout 10
bbles are situated on the table. Two players make their moves alternately. A move consists of taking
Rollout 8
since the cards are of two colors, perhaps?\n\nWait, maybe not. If each card is unique (
Rollout 13
1.2544 = 1.2544\nThen, 0.2
Rollout 8
(10,5) but considering order.\n\nWait, but when they compute the probability, is the
Rollout 14
, Take q=3/2 and similarly. Hmm. Alternatively:\n\nAssume that a=2 and
Rollout 1
as the graph of a linear operator or something. Hmm.\n\nAnother approach. Assume H has an orthon
Rollout 12
P. Turan]\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to prove that the function
Rollout 12
/ sqrt( (x^2 - a^2)(b^2 - x^2) )
Rollout 13
.2544 = 1.2544\nThen, 0.25
Rollout 10
: previous analysis says n=28:W.\n\nSo modulo7 also doesn't work.\n\nPossible idea
Rollout 13
to the nearest cent.\n<|im_start|>assistant\n<|im_start|>think\nOkay, so Joanie took a $6,
Rollout 2
\[\n x \equiv 5 \mod 8 \quad \text{and} \
Rollout 4
combination of capacitors?\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to find the percentage error
Top 16 Negative Activations
Rollout 7
is 56. Then, due to an error, all scores are increased by 5. After
Rollout 12
the variable of integration. Let me consider switching the variable to t, but not sure.\n\nWait, let
Rollout 11
+ c/a. So let's express q in terms of a, b, c. Since abc =
Rollout 5
= 181.25, greatest integer not exceeding 181.25 is
Rollout 0
Let me check. The primes less than or equal to 20 are: 2, 3
Rollout 12
, maybe we can perform a substitution in the integral to make the integral transform into another form where the dependence
Rollout 12
vanish or something.\n\nAlternatively, maybe we can substitute variables to absorb the dependence on ν?\n\nAlternatively, instead
Rollout 12
the derivative through substitution. Alternatively, consider changing the variable of integration. Let me consider switching the variable to
Rollout 3
is the molar mass. Then there's the average speed, which is sqrt(8RT/(π
Rollout 9
(10 - y)^2]\n\nSince GE = GM, then:\n\nsqrt[h^2 + y
Rollout 14
Hmmm.\n\nWait, what if we make the substitution y = sqrt(x), then dy = 1
Rollout 9
to (10,10), the altitude to EM would just be the horizontal distance from G to
Rollout 12
{-1/2} ] dt\n\nPerhaps perform substitution s = t2. Let me try:\n\nLet
Rollout 12
t2) ) ] dt\n\nLet’s do substitution t = ν sn(u, k), where sn
Rollout 12
through substitution. Alternatively, consider changing the variable of integration. Let me consider switching the variable to t,
Rollout 12
is more explicit, perhaps.\n\nLet me see. Let's try substitution.\n\nThe integral is from x =

DOWN_PROJ

Top 16 Positive Activations
Rollout 6
if the advice is generated in polynomial time, then why not just incorporate that polynomial-time computation into the main
Rollout 15
from all on, ends after two steps. Then why did I earlier thought it cycles?\n\nBecause I made
Rollout 9
confirm the positions. If the square is labeled AIME, moving around the square, then maybe A is
Rollout 7
itself isn't changed. Hmm. This is ambiguous.\n\nBut according to the problem statement: "due to
Rollout 1
find such a y by fitting it accordingly.\n\nSo why in the previous example with translated set S', it
Rollout 15
equivalent to the same system as over GF(2). Therefore if over GF(2) the system models
Rollout 12
me set t2 = ν2 + (1 - ν2) sin2θ. Then when t
Rollout 8
but since they are identical? Wait, wait, no. Wait, the problem says "shuffles the
Rollout 8
sequences. Let's consider two cases:\n\nCase 1: All 5 cards are red or all are
Rollout 12
� such that t2 = ν2 + (1 - ν2) sin2θ. Let me
Rollout 0
but a and b have exponents for the primes? Wait no, in the problem statement, the numerator
Rollout 14
the inequality holds with an excess. Hmm, then why does the inequality have coefficient8? It must be
Rollout 7
, but pass mark is fixed at 65. Wait, the scores are increased by 5,
Rollout 1
orthonormal systems can be uncountable. However, in that case, the problem still requires that
Rollout 2
for anyk is always even? No. When k=1: 9 is odd. 1
Rollout 7
increase. Hmm, so that's a critical point.\n\nWait, but reading the problem again: "due
Top 16 Negative Activations
Rollout 10
2,...\n\nDifference between 0 and 2: 2\n\n2 to 5: 3\n\n
Rollout 1
2 - 2c + d2)/2 = 0 ⇒ 0 = 0.\n\nThus
Rollout 0
}{b}\) and calculate the product of the resulting numerator and denominator. We need to determine how many
Rollout 9
2\n\nSubtract y^2 from both sides:\n\n0 = 100 - 20
Rollout 15
t+1 if and only if exactly one of its neighbors is on at minute t. But if a
Rollout 15
lamp is both ends, so does it have zero neighbors or one? Wait, the problem statement says "
Rollout 9
= ��[0 to10] [ (2/5 x +6 ) ] dx\n\n
Rollout 14
(A)))=K/A2.\n\n multiply both sides by 2 sqrt(A )) A2 )\n\nsqrt(J
Rollout 0
{a}{b}\) and calculate the product of the resulting numerator and denominator. We need to determine
Rollout 8
: 2!; total 2+2=4.\n\nCase2: for r=1,
Rollout 14
^2 ))= 1/(8*(1/(2k))*(1/k^2 ))= (
Rollout 5
that is parallel to the bases and that divides the trapezoid into two regions of equal area.
Rollout 4
recalling how to calculate the equivalent capacitance of capacitors in series and then the energy stored. \n\nFirst
Rollout 14
$f$ is integrable, and the integrals of $f(x)^2$ and $x
Rollout 1
2 = c - d2/2.\n\nSo substituting back into the previous equation:\n\n(2c
Rollout 10
initial situations in which the second player can win no matter how his opponent plays, we analyze the game where

Layer 45

GATE_PROJ

Top 16 Positive Activations
Rollout 0
, corresponding to \boxed{128}? Or wait, wait prime factors of 20
Rollout 2
the remainder is109.\n\nTherefore, answer:109.\n\nBut let me recheck all
Rollout 4
approximately 0.88%.\n\nWait, maybe 0.88 is okay. Alternatively, let
Rollout 3
168 or 169. Maybe 168 m/s.\n\nBut checking further on
Rollout 13
616\n\nTherefore, A_annual �� $9,441.12.
Rollout 5
is 181. Therefore, the answer is 181. \n\nWait, but let
Rollout 5
is 181. Therefore, final answer is 181. \n\nI think that's
Rollout 0
corresponding to \boxed{128}? Or wait, wait prime factors of 20!
Rollout 3
\boxed{170} m/s. Alternatively, could possibly round to 168 or
Rollout 4
0.88%, or maybe the answer is exact at 0.87%.\n\nBut wait,
Rollout 4
.8774964% ��0.8775%\n\nBut maybe doing
Rollout 3
\boxed{169} m/s. But need to specify.\n\nAlternatively, here's another angle
Rollout 3
68.547 m/s.\n\nSo approximately 168.5 m/s. Let
Rollout 3
Wait, but maybe the problem expects the answer as around 170 m/s. Alternatively, let
Rollout 9
\boxed{25}\n\nWait, that seems correct.\n\nLet me check this.\n\nGiven that the area
Rollout 0
So 8, so k=8. so answer is 2^(8-1)=12
Top 16 Negative Activations
Rollout 0
<|im_start|>answer\nGiven a rational number between 0 and 1, we need to write it as a
Rollout 4
_2}{C_1 + C_2}\n\]\n\nGiven \( C_1 = 2
Rollout 6
$ .\n Note that a TM $M$ with oracle access to $S$
Rollout 2
9\cdots9}_{\text{999 9's}}\) is divided by
Rollout 6
_i \) is sparse, for each length \( n \), the number of strings in \( S \
Rollout 15
at minute $t+1$ if and only if at minute $t$ , exactly
Rollout 4
Delta C_{\text{eq}}}{C_{\text{eq}}} = \frac{4
Rollout 6
\) is sparse, for each length \( n \), the number of strings in \( S \)
Rollout 14
c - d x for 0 ≤x ≤ c/d, 0 elsewhere.\n\nFirst, check that
Rollout 1
that the distance between any two distinct points in \( S \) is \( d \), we need to
Rollout 14
)\to[0,\infty)$ is integrable and $$ \int_0^\in
Rollout 9
, 0)\) to \(M(10, 10)\).\n - The apex
Rollout 2
9 \times 99 \times 999 \times \cdots \times \under
Rollout 9
}{h}x + 10 - \frac{50}{h} \right) -
Rollout 4
= \frac{4.327}{1200} \approx 0.0
Rollout 8
or a combination of one block of red and one block of green.\n\n1. **Case 1:

UP_PROJ

Top 16 Positive Activations
Rollout 7
"due to an error in the wording of a question, all scores are increased by 5. At
Rollout 9
square. Then the triangle GEM overlaps the square, and their common area is 80 square units
Rollout 7
: These are participants with original score <65, average 56. Their total score is
Rollout 7
"due to an error in the wording of a question, all scores are increased by 5. At
Rollout 7
at 65. Wait, the scores are increased by 5, so original scores become original +
Rollout 15
rule doesn't mention the current state. The next state depends only on the number of neighbors. So regardless
Rollout 7
, 6 remaining. Original average of 2 students=62, so total score=62
Rollout 7
"due to an error in the wording of a question, all scores are increased by 5. At
Rollout 8
arranged in same color order, but different actual cards, it's still counts as the same color sequence.
Rollout 2
number of terms:999. The first two terms (k=1 andk=2)
Rollout 11
if a complex number is equal to its conjugate, then it's real. So q = \over
Rollout 9
length we need to find.\n\nNow, the common area between triangle GEM and square AIME is
Rollout 7
due to an error in the wording of a question, all scores are increased by 5. At this
Rollout 8
in total. But when she lays out 5, those could be anywhere from 0 to 5
Rollout 15
of the lamp's current state, it's next state is based on the neighbors' current state.\n\nWait
Rollout 15
doesn't mention the current state. The next state depends only on the number of neighbors. So regardless of
Top 16 Negative Activations
Rollout 11
. No good.\n\nAlternatively, take a = \overline{1/b}, leading to modulus |a
Rollout 14
qualitty with K.\n\nAlternatively, start with Cauchy-Schwarz:\n\nI = ��0
Rollout 2
9\), so \(x \equiv 109 \mod 1000\).\n\n
Rollout 11
1. Not allowed. Problem explicitly says modulus not 1. So discard.\n\nPerhaps substitute three numbers as
Rollout 14
K.\n\nNo, that seems not immediately helpful. \n\nWait, have you consider applying the Cauch
Rollout 11
positive real. which is prohibited. So again invalid.\n\nThus such setup leads to real c, which is
Rollout 14
, but the second integral diverges.\n\nUnproductive.\n\nWait all right, time is passing. Let me
Rollout 1
, but assuming that S is any set. Hmm.\n\nAlternatively, since you can extend any orthonormal
Rollout 11
{1}, but c is real. No good.\n\nAlternatively, take a = \overline{1
Rollout 14
Take q=3/2 and similarly. Hmm. Alternatively:\n\nAssume that a=2 and b
Rollout 3
=168^2 + 2*168*0.5 +0.2
Rollout 6
1^log n. But still, unsure.\n\nAlternatively, this is similar to problem 1.
Rollout 10
Grundy[18}=1\n\nn=19: subtract1→18 (Grundy
Rollout 6
in S_L. However, this requires exponential time.\n\nAlternatively, assume the oracle can provide s_n in
Rollout 3
6.08=19,288.48 +199.53
Rollout 2
= 1000n + 109\), so \(x \equiv 1

DOWN_PROJ

Top 16 Positive Activations
Rollout 14
1/3} \right)^{2/ (3/2 -1)} } \right)^
Rollout 11
ate variables? Not sure. Maybe if |a| ≠ 1, but since abc = 1
Rollout 8
9 * 8 * 7 * 6. Wait, because each card is drawn without replacement
Rollout 15
)\n2. First on, second off (OF)\n3. First off, second on (FO)\n
Rollout 10
or mathematical induction.\n\nAlternatively, is there an upper limit on the number of moves? Given that each move
Rollout 3
there's a formula for the root mean square speed, which is related to temperature and molar mass.
Rollout 8
of ways to assign the red cards to the red positions and green cards to the green positions. Let me
Rollout 7
original groups and the new groups.\n\nLet me denote variables:\n\nOriginal scenario:\n\n- Number of original promoted:
Rollout 12
^{θ=π/2} dθ / t * [ sqrt( (t2 - nu2
Rollout 13
approximately $9,628.24\n\nWait, but to be precise. Let me calculate
Rollout 5
trapezoid into areas with ratio 2:3, leading us to bases of 75
Rollout 14
ponents p=2, q=4, and r=4. Let’s check:\n\n1/p +
Rollout 5
0) + (b + 100))/2 * (h/2) = (2
Rollout 7
+ x.\n\nSimilarly, new repeater count = R - x.\n\nWe are told that after the increase
Rollout 7
2)= (14/3)*(12/N)*(N)=14/3*12
Rollout 15
single lamp is both ends, so does it have zero neighbors or one? Wait, the problem statement says
Top 16 Negative Activations
Rollout 15
over GF(2). Therefore, the original problem reduces to this.\n\nHence, The answer is that
Rollout 1
is d, I need to find a point y such that { sqrt(2)/d (x -
Rollout 6
this case, the oracle TM can, given input x of length n, retrieve s_n from the oracle
Rollout 11
--> phi=0, leading to real c. Which is prohibited. Hence, unless phi=pi,
Rollout 15
(2). In such a system, the problem reduces to determining for which n the associated matrix is nil
Rollout 3
some books, they use 273 K, others 298 K. For example,
Rollout 11
1/b, would lead to c=1, which is not allowed. So perhaps variables can't be
Rollout 4
0 -60,000 +150=5,940,15
Rollout 8
color. Otherwise, if the cards were unique, it would matter which specific red or green card is where
Rollout 1
H all pairwise distance d apart. To find y such that ��2/d (x - y)
Rollout 1
y|| = d / sqrt(2),\n\nwe also have ||x - x'|| = d = sqrt
Rollout 1
∈S} is orthonormal.\n\nWhich is equivalent to requiring that { x - y } is orthogonal
Rollout 0
1 and a*b=20!. Hence, equivalent to coprime pairs (a,b) with
Rollout 4
ature with coefficients. Then the total energy error is sqrt(0.362 + (2*
Rollout 11
of a, b, c are real, which forbids this. Hence, such a case is impossible
Rollout 6
are represented in binary, then for each n and i < p(n), the length of the string would

Layer 46

GATE_PROJ

Top 16 Positive Activations
Rollout 9
Answer**\n\boxed{25}\n\n<|im_start|>answer\nSquare \(AIME\) has sides of length
Rollout 3
{169} m/s.\n\n<|im_start|>answer\nTo compute the mean molecular speed \( v \)
Rollout 11
boxed{(-1, 3)}\n\n<|im_start|>answer\nGiven three complex numbers \(a, b, c
Rollout 1
} y \text{ exists as described}}\n\n<|im_start|>answer\nGiven an infinite-dimensional Hilbert space \( \
Rollout 10
the second player can force a win.}\n\n<|im_start|>answer\nTo prove that there are infinitely many initial situations
Rollout 4
0.88\%}.\n\n<|im_start|>answer\nTo find the percentage error in the calculation of the
Rollout 11
\boxed{(-1, 3)}\n\n<|im_start|>answer\nGiven three complex numbers \(a, b,
Rollout 13
{187.12}\n\n<|im_start|>answer\nTo determine how much more money Joanie will owe
Rollout 2
**\n\boxed{109}\n\n<|im_start|>answer\nTo find the remainder when \(9 \times
Rollout 0
**\n\boxed{128}\n\n<|im_start|>answer\nGiven a rational number between 0 and 1
Rollout 1
y \text{ exists as described}}\n\n<|im_start|>answer\nGiven an infinite-dimensional Hilbert space \( \math
Rollout 5
**\n\boxed{181}\n\n<|im_start|>answer\nLet the shorter base be \( b \) and
Rollout 12
} (0, 1)}.\n\n<|im_start|>answer\nTo prove that the function \( f(\nu)
Rollout 10
second player can force a win.}\n\n<|im_start|>answer\nTo prove that there are infinitely many initial situations in
Rollout 14
) \, dx \right) }\n\n<|im_start|>answer\nTo prove the inequality \n\n\[\n\left(\
Rollout 14
(x) \, dx \right) }\n\n<|im_start|>answer\nTo prove the inequality \n\n\[\n\left
Top 16 Negative Activations
Rollout 2
P ��109 mod125.\n\nSo summary so far:\n\nP≡5 mod8
Rollout 2
k -1) �� -1 mod8.\n\nTherefore, for k=1: (2^
Rollout 0
,17,19 have exponent 1.\n\nSo primes 2: 18,
Rollout 2
)^{997} mod125\n\nBecause from k=3 to k=99
Rollout 3
^2 = 28,561. So sqrt(28,436)
Rollout 8
either order), or all reds or all greens.\n\nSo first, we can calculate the number of such
Rollout 0
= 0), so that's exponent 2.\n\nWait, 7^2 divides into 2
Rollout 8
acceptable, like RRRRR or GGGGG.\n\nSo in total, the happy sequences are those that
Rollout 4
%. So approximately 0.8776%.\n\nThus, the percentage error is approximately 0.
Rollout 4
percentage: 0.87755%. So approximately 0.8776%.\n\n
Rollout 12
] / ( (1 - nu2) nu )\n\n= - [ E(k) ] / ( (
Rollout 8
440 / 30240.\n\nSimplify this fraction.\n\nFirst, divide numerator and
Rollout 8
layout, they each need to form a single block.\n\nSo if in the laid out sequence, there is
Rollout 4
but multiplied by 2 gives 0.8%.\n\nSo, when combining these: sqrt( (0
Rollout 2
and P≡109 mod125.\n\nNow need to find a number x such that:\n\n
Rollout 2
k for k ≥3 is 0 mod8. Thus, for k ≥3, 2^

UP_PROJ

Top 16 Positive Activations
Rollout 4
0 + 3000} = \frac{6000000}{5
Rollout 4
000 + 3000} = \frac{6000000
Rollout 2
000\n \]\n\nThus, the remainder when the product is divided by 100
Rollout 9
ies h = 25\n \]\n\nThus, the length of the altitude to \(EM\)
Rollout 0
= 128\).\n\nThus, the number of rational numbers between 0 and 1 for
Rollout 2
mod 1000\n \]\n\nThus, the remainder when the product is divided by
Rollout 5
625}{2}} = \sqrt{\frac{36250}{2}} =
Rollout 9
25\n \]\n\nThus, the length of the altitude to \(EM\) in \(\
Rollout 13
187.12\n\]\n\n\[\n\boxed{187.12}\n
Rollout 5
2(b + 75) \implies 3b + 75 = 2b
Rollout 13
628.24\n\]\n\n**Difference:**\n\[\n9628.2
Rollout 9
to \(EM\) in \(\triangle GEM\) is \(\boxed{25}\).<|im_end|>
Rollout 12
0 < \nu < 1 \).\n\n\[\n\boxed{f(\nu) \text{
Rollout 2
\times 3 \times (-1)^{997} \equiv 1 \times
Rollout 4
2000 + 3000} = \frac{600000
Rollout 8
126 = 157\).\n\n\[\n\boxed{157}\n\]
Top 16 Negative Activations
Rollout 12
diverging.\n\nAlternatively, with substitution t = sinθ, but seems not applicable here. Wait, let
Rollout 12
2?\n\nAlternatively, substitution φ where t = cosθ? Let me think. For if t is from
Rollout 0
, 3^8, 5^4, etc., up to the primes less than or equal
Rollout 11
number is real, its complex conjugate is itself. So for p to be real, p = conjug
Rollout 3
the molar mass would be approximately 222 g/mol, right?\n\nSo, first, M
Rollout 11
, maybe we can write 1/a = bc, 1/b = ac, 1/c =
Rollout 11
Let’s set x = \overline{a}, y = \overline{b}, z =
Rollout 12
2 + (1 - ν2)u2), but that might complicate.\n\nAlternatively, perhaps a
Rollout 12
substitution.\n\nLet’s denote t = ν coshθ. Wait, but since t is between ν and
Rollout 12
’s do substitution t = ν sn(u, k), where sn is the Jacobi elliptic function.
Rollout 12
u) resembles an elliptic integral of the first kind. Recall that the standard elliptic integral is:\n\n
Rollout 3
the gas constant, which is 8.314 J/(mol·K).\n\nAssuming standard
Rollout 14
products. Cauchy-Schwarz comes to mind for the squared integral, but here we have three
Rollout 3
6, so atomic weight is approximately 222 (since the most stable isotope is radon
Rollout 14
{f(x)}{\sqrt{x f(x)}}}$, but that feels like a stretch and may not
Rollout 14
int x f$, so maybe split $f$ into parts such that when multiplied together, their integrals

DOWN_PROJ

Top 16 Positive Activations
Rollout 4
Let me calculate 4.327 divided by 1200.\n\nDivide by
Rollout 4
compute that more accurately.\n\n4.327 / 1200. Let me calculate
Rollout 3
=66.48*298= same approach: 66.48*
Rollout 3
6.512 * 298:\n\nLet me compute this multiplication.\n\n66.5
Rollout 4
0 / 5025. \n\nDividing 6,060,150
Rollout 4
1990 * 2985= (2000 -10)*(3
Rollout 3
66.512*293=66.512*(200
Rollout 3
2 * 298 = let's compute 66.512 * 30
Rollout 4
7 divided by 1200.\n\nDivide by 100: 4.3
Rollout 2
9= 891*999= let's compute: 891*(1
Rollout 3
6.512 * 293 = 66.512*(30
Rollout 3
.512 * 293 = 66.512*(300
Rollout 4
,150 by 5025:\n\nFirst, 5025 * 1
Rollout 13
.12 multiplied by 1.12: \n\n1.12\nx1.1
Rollout 4
. Let me calculate 4.327 divided by 1200.\n\nDivide by
Rollout 13
9441.11616 =\n\nFirst, subtracting 9441.
Top 16 Negative Activations
Rollout 14
, r=4, since 1/2 +1/4 +1/4=1.\n\n
Rollout 14
∫ f(x) dx= ��f(x)*(1 +x)^{1/2}/(
Rollout 7
, x=N/12.\n\nTherefore:\n\nA*(N/12)=14N/3
Rollout 7
2, x=1, sum=98*1=98. While student’s original score
Rollout 0
(20/27)=6 +2 +0=8\n\nFor prime 5: floor
Rollout 7
=12.\n\nx=36/12=3.\n\nOriginal repeaters:12 students
Rollout 14
)^{a} (��x f dx)^{b} }, and find a and b such that
Rollout 7
R=8.\n\nx=24/12=2.\n\nOriginal repeaters:8 students.
Rollout 7
so total score=62*2=124. Remaining 6 students average 54
Rollout 7
aters: 62*1 +54*3=62 +162=2
Rollout 14
}| dx ≤ ||f||_3 ||x^{1/3}||_3 ||x^{-
Rollout 14
+1/q +1/r =1/2 +1/4 +1/4=1.\n\n
Rollout 7
original score >=60 (since 60 +5 =65) are promoted. Therefore,
Rollout 1
each of length d / sqrt(} and whose differences have length d.\n\nBecause such vectors could form an
Rollout 9
6x ] from0 to10\n\n= (1/5)(100) +6
Rollout 9
0 to10\n\n= (1/5)(100) +60 -0 =

Layer 47

GATE_PROJ

Top 16 Positive Activations
Rollout 1
\mathcal{H}$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to show that
Rollout 11
)$ .\n\n*David Altizio*\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I have this problem here
Rollout 8
Find $m + n$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. We need
Rollout 3
Rn) in m/s\n<|im_start|>assistant\n<|im_start|>think\nOkay, I need to calculate the mean
Rollout 6
L$ . [/*]\n<|im_start|>assistant\n<|im_start|>think\nOkay, let me try to work through
Rollout 12
. [P. Turan]\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to prove that
Rollout 13
dollar value to the nearest cent.\n<|im_start|>assistant\n<|im_start|>think\nOkay, so Joanie took a $
Rollout 4
in this combination of capacitors?\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to find the
Rollout 0
!$ be the resulting product?\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I need to figure out
Rollout 9
in $\triangle GEM$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to solve this
Rollout 5
2/100$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. I need
Rollout 2
$1000$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I have this problem here
Rollout 14
) dx \right) $$ \n<|im_start|>assistant\n<|im_start|>think\nOkay, so I have to prove this
Rollout 7
oted $47$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I'm trying to solve
Rollout 10
no matter how his opponent plays.\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to prove that
Rollout 15
will be off after some time?\n<|im_start|>assistant\n<|im_start|>think\nAlright, let's try to figure out
Top 16 Negative Activations
Rollout 8
126. Then check if they can be simplified. Since 31 is prime. 1
Rollout 3
,820.576.\n\nDivide that by π * 0.222
Rollout 3
). Let's see, 168^2 = 28,224.
Rollout 8
��240= 126. So 31/126.\n\nWait
Rollout 3
28,436. Then take square root.\n\nsqrt(28,436).
Rollout 8
check by dividing numerator and denominator by 240: 7440÷240
Rollout 8
26. Then check if they can be simplified. Since 31 is prime. 12
Rollout 3
22). Wait, Radon has several isotopes, but the most common one in decay chains is
Rollout 8
Therefore, 31/126. Then check if they can be simplified. Since 3
Rollout 0
of them, 4 pairs would have a < b. Let's check. Possible assignments:\n\n1.
Rollout 0
20! but 7^3 does not.\n\nBut exponent of 7 is 2,
Rollout 0
but how many of those assignments result in a < b?\n\nSo for 720, with prime
Rollout 8
7440\n\nHence, the probability is 7440 / 302
Rollout 2
0,109. mod1000, that's109. So 8
Rollout 0
divides into 20! but 7^3 does not.\n\nBut exponent of 7 is
Rollout 2
9.\n\n9*99=891. 891 mod125:

UP_PROJ

Top 16 Positive Activations
Rollout 0
4,7:2, which are all even? Wait, 18,8,4,
Rollout 14
μ f(x) dx* λ^{-?} Wait sorry scaling variables properly:\n\nOriginal variables:\n\nLet set
Rollout 7
8, but >=60 and <65. 98 not possible.\n\nN=24
Rollout 10
n ��5k where certain conditions...} wait, unsure. Alternatively, recognize a pattern in losing
Rollout 4
000^2) )^2 )\n\nWait, this is another way of writing the terms.
Rollout 7
=65 -5 to64 -5(wait no). wait: new score >=65.
Rollout 11
1/(a2 z )/a.\n\nWait, q= a/b + b/c +c
Rollout 13
b = 0.1255\n\nBut that's not helpful. Let me compute 1
Rollout 9
/h)x + (50/h) +0? Wait, let's verify:\n\nFrom G(1
Rollout 7
score + 5 must be >=65? Wait, but if a student's original score +5
Rollout 2
=999*10 +9, wait, perhaps 9999=9*
Rollout 15
starting from O F F dies over 3 steps? Wait, in our integer simulation starting from O F
Rollout 11
. However, �� = nπ/3.\n\nSo if we check: for ��= ar
Rollout 7
(56 -54)N/3 ?\n\nWait, wait:\n\nWait 18N is
Rollout 13
55 + 255^2.\n\nBut perhaps break down using standard multiplication:\n\nBut perhaps this
Rollout 15
from O F F dies over 3 steps? Wait, in our integer simulation starting from O F F
Top 16 Negative Activations
Rollout 11
{b}{c} + \frac{c}{a}\), given that both \(p\) and
Rollout 0
}\) and calculate the product of the resulting numerator and denominator. We need to determine how many such rational
Rollout 10
- A position is losing if all possible moves lead to winning positions.\n - Through analysis, we
Rollout 15
the behavior of the system using linear algebra over the field \( \text{GF}(2) \).
Rollout 15
a lamp will be on at minute t+1 if and only if exactly one of its neighbors was on
Rollout 11
b} + \frac{b}{c} + \frac{c}{a}\), given that
Rollout 1
\in \mathcal{H} \) such that the set \( \left\{ \frac
Rollout 6
machine \( M \). By the Meyer's theorem equivalence between \( \textbf{P}/\text
Rollout 3
oltzmann distribution:\n\n\[\nv = \sqrt{\frac{8RT}{\pi M}}\n
Rollout 6
i\rangle\) and checks membership in \( S \).\n\n---\n\n**Problem 2:**\n\nFor \
Rollout 1
in \mathcal{H} \) such that the set \( \left\{ \frac{\
Rollout 1
d, the resulting set is an orthonormal system. That means each vector has length 1 and
Rollout 6
this string, and accepts if the oracle confirms its presence.\n\n**Answer:** \nThe sparse set \( S
Rollout 15
's next state is determined by the XOR (sum modulo 2) of its neighbors' states.\n\n1
Rollout 12
this integral as a form of the complete elliptic integral of the first kind, \( K(k) \
Rollout 1
right\} \) is an orthonormal system.\n\n1. **Orthonormal Conditions**:\n

DOWN_PROJ

Top 16 Positive Activations
Rollout 4
0.52 +0.52 +something)%? Wait, no—wait:\n\nWait,
Rollout 13
^2.\n\nPerhaps set a = 1.25, b=0.0044
Rollout 12
{-1/2} ] dt\n\nPerhaps perform substitution s = t2. Let me try:\n\nLet
Rollout 14
f(x) dx ≤2.\n\nIf I can show that, then returning to general J and K,
Rollout 4
pF. That is 4.326/1200≈0.00
Rollout 14
�f(x) dx ≤2.\n\nIf I can show that, then returning to general J and K
Rollout 12
2 ). Hmm, this is not the standard form. However, perhaps with substitution.\n\nAlternatively, substitute phi
Rollout 7
=98 is impossible.\n\nTherefore, inconsistency. What does that mean? It means that our assumptions might
Rollout 10
18 is a N-position, which it may be, as haven't analyzed. Let's check
Rollout 12
this helps in differentiation.\n\nAlternatively, another substitution. Consider that in the integrand, sqrt( (t
Rollout 4
C_eq was 4.327 pF, C_eq is 1200 p
Rollout 2
mod8.\n\nTherefore, equation becomes:\n\n(5m +5)≡5 mod8\n\nSubtract
Rollout 15
2) model is insufficiently modeling the actual system. Hence, perhaps equivalence is not correct.\n\nBut why
Rollout 7
score + 5 must be >=65? Wait, but if a student's original score +5
Rollout 3
)^2=168^2 + 2*168*0.5 +0
Rollout 1
for all x, x' in S?\n\nSuppose that all points in S lie on a sphere in
Top 16 Negative Activations
Rollout 0
2,3,5:\n\nThe 8 assignments:\n\n1. Assign none to a => a=1
Rollout 13
1.12\nx1.12\n--------\n224 (1.12 *
Rollout 3
, I need to recall. In some problems, if the temperature isn't given, it's assumed to
Rollout 13
55*0.1255. Wait, maybe better to compute directly.\n\nAlternatively:\n\n(
Rollout 13
(the last digit after decimal of the second number):\n\n1.1255 * 0.
Rollout 11
, taking reciprocal vectors.\n\nAlternatively, this can't take too long, given that it's an Olympiad
Rollout 15
neighbor. For the middle lamp L2, it has two neighbors.\n\nLet me consider some example transitions.
Rollout 13
.00551936.\n\nHmm, sorry, not the best approach. Wait,
Rollout 4
00 ± 0.02 V. Hmm, alright, let's start by recalling how to
Rollout 15
its neighbors is on (the two lamps at the ends have one neighbor each, all other lamps have two
Rollout 13
interest compounds quarterly compared to if it compounds annually. Hmm, I need to calculate the amount owed in both
Rollout 13
place (tenths, hundredths, etc.)\nWait, the multiplicand is 1.12
Rollout 14
/12)^{1/3}=12^{-1/3}.\n\nThus, μ=
Rollout 6
I'm going in circles here. Given the time constraints, maybe for problem 3, since it's
Rollout 4
5) is sqrt(10^-5) = 10^-2.5 ��
Rollout 13
't making any payments for 4 years, and then she has to pay back everything she owes at the

Layer 48

GATE_PROJ

Top 16 Positive Activations
Rollout 14
) \, dx \right) }\n\n<|im_start|>answer\nTo prove the inequality \n\n\[\n\left(\
Rollout 9
Answer**\n\boxed{25}\n\n<|im_start|>answer\nSquare \(AIME\) has sides of length
Rollout 0
**\n\boxed{128}\n\n<|im_start|>answer\nGiven a rational number between 0 and 1
Rollout 1
y \text{ exists as described}}\n\n<|im_start|>answer\nGiven an infinite-dimensional Hilbert space \( \math
Rollout 2
**\n\boxed{109}\n\n<|im_start|>answer\nTo find the remainder when \(9 \times
Rollout 12
} (0, 1)}.\n\n<|im_start|>answer\nTo prove that the function \( f(\nu)
Rollout 11
boxed{(-1, 3)}\n\n<|im_start|>answer\nGiven three complex numbers \(a, b, c
Rollout 4
0.88\%}.\n\n<|im_start|>answer\nTo find the percentage error in the calculation of the
Rollout 8
**\n\boxed{157}\n\n<|im_start|>answer\nKathy has 5 red cards and 5
Rollout 5
**\n\boxed{181}\n\n<|im_start|>answer\nLet the shorter base be \( b \) and
Rollout 10
second player can force a win.}\n\n<|im_start|>answer\nTo prove that there are infinitely many initial situations in
Rollout 13
{187.12}\n\n<|im_start|>answer\nTo determine how much more money Joanie will owe
Rollout 3
{169} m/s.\n\n<|im_start|>answer\nTo compute the mean molecular speed \( v \)
Rollout 14
(x) dx \right) $$ \n<|im_start|>assistant\n<|im_start|>think\nOkay, so I have to prove
Rollout 15
for non-negative integers \( k \).\n\n<|im_start|>answer\nTo determine for which values of \( n \)
Rollout 10
win no matter how his opponent plays.\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to prove
Top 16 Negative Activations
Rollout 11
variables is the product of conjugates of the others. For instance, maybe c = \overline{
Rollout 4
49e-3*sqrt(10). Because sqrt(10)=3.16
Rollout 4
Wait, this is another way of writing the terms. Maybe this will increase precision?\n\nWait, or rather
Rollout 13
5. Alternatively, perhaps use the standard multiplication grid. But this might take longer.\n\nAlternatively, another method
Rollout 11
y/z, since z is conjugate of c. Hmmm, perhaps requiring relationships like a/b =
Rollout 10
), so subtracting squares could alter the parity differently.\n\nAlternatively, consider breaking the game into two parts:
Rollout 12
Alternatively, substitution φ where t = cosθ? Let me think. For if t is from ν to
Rollout 1
property) and maybe in a weakly compact set. But in Hilbert spaces, closed balls are weak
Rollout 6
to use the same number of bits for each n. Wait, but log n is variable length. If
Rollout 14
some of the integrals become 1 to normalize. Suppose we set J_λ=1 and K
Rollout 12
dt\n\nPerhaps perform substitution s = t2. Let me try:\n\nLet s = t2, so
Rollout 13
, that's too big. Alternatively, use written multiplication.\n\nLine up 1.1255
Rollout 14
f, such that when multiplied with some other functions...\n\nAlternatively, use a substitution t= x. But
Rollout 14
-Schwarz on $I$? Let's see.\n\nBut $I$ is the integral of $
Rollout 11
maybe multiply q by some term and combine with p. For example, if q = a/b + b
Rollout 12
?\n\nAlternatively, substitution φ where t = cosθ? Let me think. For if t is from ν

UP_PROJ

Top 16 Positive Activations
Rollout 0
a share no primes with b. Therefore, the coprime pairs (a,b) with a*b
Rollout 15
4. Therefore, according to the linear algebra model, any initial state should die out after at most
Rollout 13
is 187.12 which seems reasonable.\n\nSo my conclusion is that the extra amount Jo
Rollout 0
, both positive integers, a < b.\n\nAnd this number is equal to 2^{k -1
Rollout 5
= 181.25, greatest integer not exceeding 181.25 is
Rollout 5
181.25, greatest integer not exceeding 181.25 is 1
Rollout 3
/(πM)). The problem mentions "mean molecular speed," which is usually the average speed. Wait,
Rollout 5
5 is 181. Therefore, final answer is 181. \n\nI think that
Rollout 0
20! and a < b.\n\nThus previous logic applies; each such pair is obtained by assignment of
Rollout 0
\(a\) and \(b\) are coprime factors of 20!. So they form a
Rollout 0
=N!:\n\nSince a and b are coprime.\n\nHence equivalence between choices over subsets of primes
Rollout 10
2 is not in S. Therefore, such a set S implies that every number in S is a losing
Rollout 10
The problem states there are infinitely many such positions, so we have to prove that.\n\nFirst, let's
Rollout 15
index 4. Therefore, according to the linear algebra model, any initial state should die out after at
Rollout 0
\(a = b\). So all coprime pairs with \(a \times b = 2
Rollout 0
when we write a and b as coprime factors, the way I'm thinking is that the assignment
Top 16 Negative Activations
Rollout 7
= (56 -54)N/3 ?\n\nWait, wait:\n\nWait 18N
Rollout 11
a bc =1, so ab*c =a. So a b c =1. Hence, multiplying
Rollout 13
, b = 0.1255\n\nBut that's not helpful. Let me compute
Rollout 7
P =79x - (A +5)x\n\n-3P=(74 -A)
Rollout 14
\( x f(x) \geq A f(x) \), so:\n\n\[\nI_2
Rollout 14
} )^{.1 } + sqrt(J)/2.\n\n Wait, actual calculation:\n\nTerm1: sqrt
Rollout 11
/ sinθ - r3 -1/r3 ] =0\n\nWait, let's parse:\n\nE
Rollout 2
2^k \equiv 0 \mod 8\), so \(2^k - 1
Rollout 11
b|^2*(|b|^4 a ) ) = |b|^6 a → Therefore, |
Rollout 14
sqrt{A} + \frac{K}{A}.\n\]\n\nWe then minimize the right-hand side
Rollout 14
4} + ( ��J )) /2.\n\n Well, no, the expression is(√
Rollout 4
}{(C_1 + C_2)^2} \quad \text{and} \quad
Rollout 14
�A^∞ f(x) *x /x dx ≤ (��A^∞ (f(x
Rollout 14
sqrt{J}} \right)^{2/3}}.\n\]\n\nSimplifying both terms,
Rollout 12
\nu^2} \cdot \nu^2}\n\]\n\nSimplifying this expression, we
Rollout 7
213N -209N)/3=4N/3.\n\nTherefore:\n\n16

DOWN_PROJ

Top 16 Positive Activations
Rollout 2
$1000$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I have this problem here
Rollout 8
Find $m + n$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. We need
Rollout 7
oted $47$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I'm trying to solve
Rollout 5
2/100$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. I need
Rollout 9
in $\triangle GEM$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to solve this
Rollout 10
no matter how his opponent plays.\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to prove that
Rollout 15
will be off after some time?\n<|im_start|>assistant\n<|im_start|>think\nAlright, let's try to figure out
Rollout 11
)$ .\n\n*David Altizio*\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I have this problem here
Rollout 6
L$ . [/*]\n<|im_start|>assistant\n<|im_start|>think\nOkay, let me try to work through
Rollout 4
in this combination of capacitors?\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to find the
Rollout 13
dollar value to the nearest cent.\n<|im_start|>assistant\n<|im_start|>think\nOkay, so Joanie took a $
Rollout 0
!$ be the resulting product?\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I need to figure out
Rollout 3
Rn) in m/s\n<|im_start|>assistant\n<|im_start|>think\nOkay, I need to calculate the mean
Rollout 1
\mathcal{H}$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to show that
Rollout 12
. [P. Turan]\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to prove that
Rollout 3
temperature is given, perhaps they just want the formula, but I don't think so. Wait, if
Top 16 Negative Activations
Rollout 15
whether that single neighbor is on. For all other lamps, since they have two neighbors, their next state
Rollout 2
9= 891*999= let's compute: 891*(1
Rollout 2
9: 891*999=891*(1000 -1
Rollout 14
$-\mu K$ is $-\mu x$. So setting the functional derivative to zero gives:\n\n$$
Rollout 14
dx - d \int x^2 dx = c*( ( (c/d)^2 ) /2
Rollout 3
6.512 * 298 = ?\n\nAs before:\n\n66.512
Rollout 11
overline{a} b = |b|^2*(|b|^4 a ) ) = |
Rollout 4
5). Well, sqrt(7.7) is approx 2.7749, so
Rollout 7
5 -59)x=16x.\n\nTherefore, the equation becomes:\n\n209N/
Rollout 0
so that's included.\n\nBut if a and b can be written as a = 1 and b =
Rollout 8
she has 5 red cards and 5 green cards, with each color being indistinct, then
Rollout 15
maybe when n=1, it's a line with a single lamp. In that case, since it
Rollout 13
51936*6000 = 21.11616\n\n
Rollout 7
75 -59)x=16x.\n\nTherefore, the equation becomes:\n\n209N
Rollout 3
66.512*293=66.512*(200
Rollout 14
-\mu K$ is $-\mu x$. So setting the functional derivative to zero gives:\n\n$$\n

Layer 49

GATE_PROJ

Top 16 Positive Activations
Rollout 6
then O(n^k / log n), but this is still not polynomial per length. Wait.\n\nAlternatively
Rollout 6
have p(n) strings for each n, but this might not be sparse if p(n) is a
Rollout 15
which is 2^2 -1: no, since n=3 was bad. Maybe another pattern
Rollout 10
: subtract16=10 (L) which is previous, so from 26 can go
Rollout 11
0. Then p= -2. But maybe this is invalid.\n\nGiven previous problem maybe David Altizio
Rollout 15
. All off. So within one step! But this seems contradictory. Wait, maybe a mistake.\n\nWait
Rollout 2
that 109 mod8=5, which satisfies the first congruence.\n\nWait, hmm
Rollout 2
109 mod8 is 5, which matches. So 109 is indeed congr
Rollout 15
they die in five steps. Similarly, other starting states can cycle.\n\nWait, but in GF(2
Rollout 10
moves: take 4, leave 1, which is winning. Hmm, wait. So if from
Rollout 11
Then p= -2. But maybe this is invalid.\n\nGiven previous problem maybe David Altizio often uses
Rollout 6
their values. Wait, that was the prior idea which leads to non-sparsity.\n\nAlternatively, define S
Rollout 6
an index (n, i). But again, this requires handling indices, and similar length as before.\n\n
Rollout 2
: 109 mod8=5, which matches. Therefore, the solution is x≡1
Rollout 14
(∫x^{2/3} dx ). Again, but the second integral diverges.\n\nUnproductive
Rollout 14
/q=1. So q=2. But again, integrating 1 over �� is infinite.\n\n
Top 16 Negative Activations
Rollout 3
(Rn) in m/s\n<|im_start|>assistant\n<|im_start|>think\nOkay, I need to calculate the
Rollout 4
stored in this combination of capacitors?\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to find
Rollout 3
) in m/s\n<|im_start|>assistant\n<|im_start|>think\nOkay, I need to calculate the mean molecular speed
Rollout 11
,q)$ .\n\n*David Altizio*\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I have this problem
Rollout 3
on (Rn) in m/s\n<|im_start|>assistant\n<|im_start|>think\nOkay, I need to calculate
Rollout 14
(x) dx \right) $$ \n<|im_start|>assistant\n<|im_start|>think\nOkay, so I have to prove
Rollout 1
\mathcal{H}$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to show that
Rollout 3
Rn) in m/s\n<|im_start|>assistant\n<|im_start|>think\nOkay, I need to calculate the mean
Rollout 13
a dollar value to the nearest cent.\n<|im_start|>assistant\n<|im_start|>think\nOkay, so Joanie took a
Rollout 12
$ . [P. Turan]\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to prove
Rollout 4
energy stored in this combination of capacitors?\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to
Rollout 1
$ \mathcal{H}$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to show
Rollout 13
dollar value to the nearest cent.\n<|im_start|>assistant\n<|im_start|>think\nOkay, so Joanie took a $
Rollout 6
$L$ . [/*]\n<|im_start|>assistant\n<|im_start|>think\nOkay, let me try to
Rollout 2
$1000$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I have this problem here
Rollout 4
in this combination of capacitors?\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to find the

UP_PROJ

Top 16 Positive Activations
Rollout 6
S is sparse.\n\nThen, the oracle machine M, when given <x,i>, encodes i and
Rollout 6
bit of α_n. Then, the machine M, given x of length n, needs to get the
Rollout 14
.\n\nFor I1=∫0^A f dx:\n\n applyCauchy-Schwarz: I
Rollout 6
sparse.\n\nTo retrieve α_n, the oracle machine, given input x of length n, would need to
Rollout 0
how many rational numbers between 0 and 1", each such number is associated with a unique pair (
Rollout 6
S_i, then S is sparse. Then, when given <x, i>, the machine encodes
Rollout 1
0.\n\nCalculating the norm of v_x: ||v_x|| = (sqrt(2)/
Rollout 6
Thus, S is sparse.\n\nThen, the oracle machine M, when given <x,i>, encodes
Rollout 6
in S_i, then S is sparse. Then, when given <x, i>, the machine enc
Rollout 6
is sparse.\n\nThen, the oracle machine M, when given <x,i>, encodes i and x
Rollout 3
25°C for average speeds.\n\nHence, if following such an example, this question's answer would
Rollout 0
\(a \times b = 20!\), then each of the prime powers in the factorization
Rollout 6
, S is sparse.\n\nThen, the oracle machine M, when given <x,i>, encodes i
Rollout 6
Then, when given <x, i>, the machine encodes (i,x) into a string s
Rollout 9
Wait no, if the altitude is from G to EM, which is length h, then if h is
Rollout 12
/ sqrt(1 - k2 cos2theta) = 1 / sqrt(1 - k2
Top 16 Negative Activations
Rollout 4
} = \sqrt{12.96 + 5.76} = \sqrt{
Rollout 5
3b + 75 = 2b + 150 \implies b =
Rollout 7
=486. Total=186+486=672=56
Rollout 3
so 19,953.6 - 465.584 =1
Rollout 3
pi*0.028= ~0.08796\n\n19,8
Rollout 7
05N/3=(213N -205N)/3=8N/
Rollout 4
42)=sqrt(12.96+5.76)=sqrt(18.
Rollout 3
*0.028= ~0.08796\n\n19,82
Rollout 4
42)=sqrt(12.96+5.76)=sqrt(18.
Rollout 3
, so ~7,482.6 - 49.884 = 7
Rollout 5
3b + 75 = 2b + 150 \implies b =
Rollout 8
5*4*3)=2*20*60=2400\n\nk=
Rollout 2
solution is x≡109 mod1000.\n\nWait, but hold on, earlier we
Rollout 4
00013 + 0.000064) �� sqrt(0
Rollout 5
3b + 75 = 2b + 150; subtract 2b
Rollout 10
2 modulo 2 and 5 modulo 5? Not sure. Alternatively, these numbers could be written

DOWN_PROJ

Top 16 Positive Activations
Rollout 4
3*25.2004=603*25 +603*
Rollout 4
004e-12 �� 603*25.2004
Rollout 13
Adding them together: 1.0609 + 0.063654
Rollout 8
denominator by 240: numerator becomes 7440 �� 240=
Rollout 15
from just L1 on it cycles. Therefore, same conclusion: n=3 is invalid since some initial
Rollout 13
\n\nSo combining all:\n\n1.5625 + 0.011 + 0
Rollout 5
zoid is $100$ units longer than the other base. The segment that joins the mid
Rollout 0
fraction \(\frac{a}{b}\) where \(0 < a < b\) and \(a
Rollout 6
. In this case, since the angel string is fixed for each input length, and computed deterministically,
Rollout 8
= 31 +126= 157.\n\n**Final Answer**\n\boxed{
Rollout 0
rational number between 0 and 1 can be written as a fraction \(\frac{a}{b
Rollout 11
/(r2 e^{i2θ}) would be similar to 2x cosφ if 2
Rollout 10
1→14≡4 mod5, which is a winning position. Subtract4→11
Rollout 9
10 units. But in the square, that side EM is part of the square. Then the
Rollout 1
∈ S } are orthonormal. Hence, they can be considered as an orthonormal basis for
Rollout 1
non-empty.\n\nBut in infinite-dimensional spaces, you can have decreasing sequences of closed convex sets with empty intersection
Top 16 Negative Activations
Rollout 7
5)x\n\n-3P=(74 -A) x.\n\nWait, substitute P=2N
Rollout 7
/3) = (75*2)/3 *N=150/3 *N
Rollout 7
this math problem about participants in a test. Let me first try to understand what's being asked here.\n\n
Rollout 7
)x\n\n-3P=(74 -A) x.\n\nWait, substitute P=2N/
Rollout 11
8) sin3θ = (63/8) sin3θ\n\nFor q to be real
Rollout 1
2 - (2α - ||y||2) = 0 ⇒ (2c - d2
Rollout 9
: GE: y=(-1/5)(10)+2= -2 +2=0,
Rollout 1
)/2 - (2α - ||y||2) = 0 ⇒ (2c - d
Rollout 7
79P =79x - (A +5)x\n\n-3P=(74 -
Rollout 9
00 => y = 5.\n\nTherefore, the y-coordinate of point G must be 5,
Rollout 1
4 = d2 - (6d2/4) = d2 - 3)1.
Rollout 7
, the x terms:\n\n(75 -59)x=16x.\n\nTherefore, the equation
Rollout 7
+59*(N/3) -59x =71N\n\nCalculate the terms:\n\n
Rollout 8
5−r greens in order where all reds are adjacent and all greens adjacent. So the number of
Rollout 7
score becomes 66N + 5N = 71N. But also, the total
Rollout 11
) sin3θ = (63/8) sin3θ\n\nFor q to be real,

Layer 50

GATE_PROJ

Top 16 Positive Activations
Rollout 4
. The equivalent capacitance for capacitors in series is given by:\n\n\[\nC_{\text{
Rollout 2
times (-1)^{997} \equiv 1 \times 3 \times (-1
Rollout 14
/2 +1/4 +1/4=1.\n\nThen Hölder gives:\n\nI ≤ ||
Rollout 9
\(x = 0\) to \(x = 10\):\n \[\n \text
Rollout 14
/2 +1/4 +1/4=1.\n\nThus, Hölder gives:\n\nI =
Rollout 8
400 + 1200 = 7200\).\n\nAdding both cases,
Rollout 5
} = \frac{2}{3}\n\]\nSolving this, we find:\n\[\n3
Rollout 8
= 7440\).\n\nThe probability is \(\frac{7440}{3
Rollout 5
midpoints of the legs, has a length equal to the average of the two bases, \( b +
Rollout 9
0}{h} = 80 \implies \frac{500}{h} =
Rollout 8
same color).\n - Total for Case 1: \(2 \times 5! = 2
Rollout 12
nu) = -\frac{E(\sqrt{1 - \nu^2}) - \nu
Rollout 2
^1 - 1)(2^2 - 1) \times (-1)^{99
Rollout 13
1.2544^2 \approx 1.57351936
Rollout 7
. Total=186+486=672=56*12.
Rollout 9
500}{h}\n \]\n - Setting the area equal to 80:\n
Top 16 Negative Activations
Rollout 3
161.4 m/s.\n\nAs an assistant, I need to perhaps suggest both possibilities, but
Rollout 3
m/s. But need to specify.\n\nAlternatively, here's another angle: gas thermometric speeds. At
Rollout 11
some specific curves? Not sure.\n\nWait, another thought. Since abc = 1, take logarithms
Rollout 8
400 + 2400 + 1200 = (1200
Rollout 11
of the other two. But maybe no.\n\nAnother thought: If both p and q are real, maybe
Rollout 11
: (a b)^3\n\nTerm5: 1/a3\n\nTerm6: 1/(
Rollout 12
^2 - ν^2)(1 - t^2) ) ] dt\n\nThen, integrating from
Rollout 4
) method.\n\nAlternatively, check the question one more time. The question says: "What is the percentage
Rollout 1
imposed by each x in S are consistent.\n\nAnother thought: think of 2y as a functional.
Rollout 1
al through such a shift and scaling.\n\nSumming up, then despite the difficulty with coordinate system, using
Rollout 6
system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet's say a language $L \
Rollout 8
system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nKathy has $5$ red cards and
Rollout 7
system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nIn a mathematics test number of participants is
Rollout 10
system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nConsider the following two person game. A number
Rollout 9
system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nSquare $AIME$ has sides of length
Rollout 11
system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nSuppose $a,\,b,$

UP_PROJ

Top 16 Positive Activations
Rollout 1
x1 - x2, 2y⟩ = ||x1||2 - ||x2||
Rollout 11
)( r e^{iθ}) ) = r2 + (1/r) e^{-iθ}
Rollout 1
x and x',\n\n||x - x'||2 = d2,\n\nand ⟨x, x'⟩
Rollout 2
,000 -891=890,109. mod100
Rollout 2
9*9999≡109*999≡109* -1
Rollout 1
d, then:\n\n||x - x'||2 = d2,\n\nwhich is ⟨x - x',
Rollout 1
.\n\nBut using the inner product formula:\n\nd2 = ⟨x - x', x - x'⟩
Rollout 2
00 -891=890,109. mod1000,
Rollout 1
we can check:\n\n||x - x'||2 = ||(x - y) - (x'
Rollout 2
1). But (10^4 -1)=9999. 9999
Rollout 2
9\n\nk=3, so10^3 -1=999≡-1 mod
Rollout 9
ies \frac{500}{h} = 20 \implies h = 2
Rollout 2
9: 891*999=891*(1000 -1
Rollout 2
000 -109=108,891≡891 mod1
Rollout 1
S, then ⟨x - x', y⟩ = ⟨x, y⟩ - ⟨x', y
Rollout 14
I2: ��A^∞ f(x) dx = ��A^∞ f(x)
Top 16 Negative Activations
Rollout 4
4)/2≈6 pF. Wait, but when we did the partial derivatives, we got ~
Rollout 0
8\), since each pair is counted twice. But actually, since we are considering ordered pairs where a
Rollout 3
people use 25°C as standard conditions. But in the absence of information, I think STP
Rollout 8
,5) if we have an unordered hand. However, since the order matters, the number of sequences
Rollout 7
suggests that perhaps the increase affects the promotion status? But the problem statement seems a bit ambiguous here. If
Rollout 4
is [0.003]^2. But wait:\n\nFirst term inside the formula is [ (
Rollout 9
and x=10 have different distance between. But in reality, these lines are not vertical but the
Rollout 13
higher than 1.5625.\n\nBut the exact calculation gives around 1.57
Rollout 15
maybe n=3 isn't cyclic.\n\nWait, but earlier thinking process stated there's a cycle, but
Rollout 9
the diagonal from E to M? Wait, no. Wait, E is bottom-right and M is top
Rollout 9
h is from 0 to 10? But EM is the vertical edge of the square. So
Rollout 3
a bit different from the root mean square speed. But maybe in this context, they just want the root
Rollout 4
0.003\n\nWait, no, actually. The term inside the sqrt for the first
Rollout 0
0! but 7^3 does not.\n\nBut exponent of 7 is 2, which is
Rollout 4
6, 2.6% total spread. But since normally uncertainty is defined as ± half the total
Rollout 9
=0 and x=10). Wait, but in trapezoid area formula:\n\nAverage of

DOWN_PROJ

Top 16 Positive Activations
Rollout 12
) = K(k), with dk/dnu = ( -nu / sqrt(1 - nu^2
Rollout 12
u) = K(k), with dk/dnu = ( -nu / sqrt(1 - nu^
Rollout 3
for Rn, which is approx. 1/8 the speed (mass is ~8 times higher
Rollout 3
5.87 /337≈0.047.\n\nTherefore, sqrt≈16
Rollout 2
1mod125. 125*7=875, 891
Rollout 6
polynomial time given the input length (i.e., 1^n), then the advice isn't giving any
Rollout 3
5.87/337≈0.047. Hence, 168
Rollout 3
5.87)/337≈0.047. Thus, sqrt≈16
Rollout 15
state of the lamps changes: A lamp is on at minute $t+1$ if and
Rollout 2
1 mod125: 125*7=875. 891
Rollout 12
/dk = ��0^{pi/2} [ (k sin2 theta ) / (1
Rollout 12
2, so t = sqrt(s), dt = (1/(2 sqrt(s))) ds.\n\nThen,
Rollout 3
)=15.87/337≈0.047. Hence, 1
Rollout 9
For h=20, point G is at 10 -20 = -10,
Rollout 6
, but the problem doesn't say that k is given as part of the input. Wait, the input
Rollout 6
advice α_n. But encoding n can be done via binary representation. Then for each n, the number
Top 16 Negative Activations
Rollout 15
in our first example with n=4 (all lamps on), the system actually reaches all off in
Rollout 12
), therefore, f’(nu) = dK/dk * dk/dnu\n\n= [ E
Rollout 7
7)there is no solution. But the problem says "Find all possible values of N", so maybe
Rollout 3
temperature is required? But no, the problem says "compute the mean molecular speed," so they must want
Rollout 15
, in our n=4 initial calculation with all lamps on, the system died after five steps. However
Rollout 12
’(nu) = dK/dk * dk/dnu\n\n= [ E(k) - (
Rollout 12
(k) >0. But the numerator in dK/dk is this expression, but wait in d
Rollout 8
to count how many of these sequences have all red cards adjacent and all green cards adjacent.\n\nSo possible scenarios
Rollout 3
the temperature is required? But no, the problem says "compute the mean molecular speed," so they must
Rollout 7
But that's conflicting with the problem's request to "Find all possible values...", unless there are constraints I
Rollout 4
55%. So approximately 0.8776%.\n\nThus, the percentage error is approximately
Rollout 12
) / (1 - ν2 )\n\nThus, sinθ cosθ = sqrt( (t2 -
Rollout 12
/ (k nu2 )\n\nThen, multiplying by dk/dnu = - nu /k:\n\nf’
Rollout 14
^{1/(2a -1)}}\n\nFrom K_λ=1: λ^{a -2
Rollout 12
dt / sqrt(...) becomes dθ / t\n\nBut t is expressed in terms of theta via t =
Rollout 12
.\n\nThus, f’(nu) = dK/dk * dk/dnu.\n\nI need to

Layer 51

GATE_PROJ

Top 16 Positive Activations
Rollout 5
+ x)) = 125H\n\nLet me factor H out:\n\n(x + 17
Rollout 5
175)/(x + 75)\n\nLet me write the entire equation:\n\n(x + 1
Rollout 11
i2θ}))(re^{iθ})\n\nLet me compute step by step:\n\n- a + b
Rollout 11
2θ}))(re^{iθ})\n\nLet me compute step by step:\n\n- a + b =
Rollout 5
75) = 125\n\nLet me let y = x + 75.
Rollout 5
x)) = 125H\n\nLet me factor H out:\n\n(x + 175
Rollout 9
, maybe visualizing coordinates would help. Let's place EM as the vertical segment from (10,
Rollout 4
15 )^2\n\nHmm, let's compute these step by step.\n\nFirst, compute the
Rollout 11
i sin2θ) = 0\n\nDivide by 2i:\n\n3 sinθ + (
Rollout 9
,10). So these are straight lines. Let me find equations for GE and GM.\n\nEquation
Rollout 11
Similarly, simplify: + xy + terms. Let me compute each term:\n\nFirst, p = x +
Rollout 13
1.2544 squared. Let's break it down:\n\nFirst, 1 *
Rollout 13
.12 multiplied by 1.12: \n\n1.12\nx1.1
Rollout 12
)(1 - t^2) )\n\nHmm. Let's compute the expression under the square root:\n\nFirst
Rollout 5
75)/(x + 75)\n\nLet me write the entire equation:\n\n(x + 17
Rollout 5
b = 175, height H, area S = (75 + 175
Top 16 Negative Activations
Rollout 4
energy stored in this combination of capacitors?\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to
Rollout 11
\boxed{(-1, 3)}\n\n<|im_start|>answer\nGiven three complex numbers \(a, b,
Rollout 2
is \(\boxed{109}\).<|im_end|>
Rollout 9
\) is \(\boxed{25}\).<|im_end|>
Rollout 10
no matter how his opponent plays.\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to prove that
Rollout 14
) \, dx \right) }\n\]<|im_end|>
Rollout 0
!$ be the resulting product?\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I need to figure out
Rollout 12
$ . [P. Turan]\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to prove
Rollout 15
\]\n\nfor non-negative integers \( k \).<|im_end|>
Rollout 0
is \(\boxed{128}\).<|im_end|>
Rollout 3
(Rn) in m/s\n<|im_start|>assistant\n<|im_start|>think\nOkay, I need to calculate the
Rollout 9
in $\triangle GEM$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to solve this
Rollout 3
on (Rn) in m/s\n<|im_start|>assistant\n<|im_start|>think\nOkay, I need to calculate
Rollout 9
EM$ in $\triangle GEM$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to
Rollout 8
Find $m + n$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. We need
Rollout 10
can win no matter how his opponent plays.\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to

UP_PROJ

Top 16 Positive Activations
Rollout 5
exceeding x2/100. \n\nFirst, let me recall some trapezoid properties.
Rollout 10
2a + 3b, but that's not helpful. Wait, 0 is even,
Rollout 3
molecular speed of radon gas. Hmm, first, I remember there's a formula for the root mean
Rollout 15
can be such that all initial states die out. But maybe not?\n\nAlternatively, perhaps the good ns are
Rollout 1
cal{H}$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to show that there exists
Rollout 14
\right) $$ \n<|im_start|>assistant\n<|im_start|>think\nOkay, so I have to prove this inequality:
Rollout 1
/ sqrt{1 - a^, but this is stretching.\n\nBut since in each pair x, x
Rollout 7
$47$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I'm trying to solve this math
Rollout 13
5 by 11255.\n\nThis is equal to (11000 +
Rollout 14
c/(1 + d x)$, but that's a guess.\n\nBut this process is getting a bit
Rollout 11
*David Altizio*\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I have this problem here where a
Rollout 12
)(t + b)} } \n\nBut maybe that's diverging.\n\nAlternatively, with substitution t = sin
Rollout 14
d x)$, but that's a guess.\n\nBut this process is getting a bit convoluted.
Rollout 5
100$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. I need to solve
Rollout 8
m + n$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. We need to find
Rollout 15
off after some time?\n<|im_start|>assistant\n<|im_start|>think\nAlright, let's try to figure out this problem
Top 16 Negative Activations
Rollout 14
] = ��f dx ~ [f] * [x],\n- [J] = ��
Rollout 3
s2·mol). Then, dividing by M in kg/mol, the mol unit cancels, then
Rollout 14
the integrals:\n\n- [I] = ��f dx ~ [f] * [x],\n
Rollout 3
*R*T)/(π*M)), R is in J/(mol·K), which is same as (kg
Rollout 3
m2)/(s2·mol) divided by (kg/mol) is (m2/s2).
Rollout 3
J/(mol·K), which is same as (kg·m2)/(s2·mol·
Rollout 11
^{iθ} ] + (1/r3)[ e^{-iθ} -e^{iθ
Rollout 14
- [I] = ��f dx ~ [f] * [x],\n- [J]
Rollout 14
:\n\n- [I] = ��f dx ~ [f] * [x],\n- [J
Rollout 3
·m2)/(s2·mol) divided by (kg/mol) is (m2/s2
Rollout 8
the probability is 31/126, so m=31, n=12
Rollout 8
). But the way the problem is phrased: "all the red cards laid out are adjacent and
Rollout 8
m=31, n=126.\n\nWait, but let's confirm. Wait, when
Rollout 7
12:\n\nA -42=56\n\nThus,A=98\n\nWait, but A
Rollout 7
b) impossible, no solutions. But language problems say "Find all possible values ofN". If "
Rollout 7
24,36. However, the problem states N<40. 36 is less

DOWN_PROJ

Top 16 Positive Activations
Rollout 13
) = 11000^2 + 2*11000*2
Rollout 0
), it depends. As with N! for N>1.\n\nWait, for N >=2, N
Rollout 5
. Probably, I made an algebraic mistake.\n\nStarting from the two equations:\n\n1. (75
Rollout 10
k2 and (k+1)2 is 2k+1, which is greater than k
Rollout 0
0 and 1 with product a*b=720, gcd(a,b)=1, a <
Rollout 1
, since each sphere is closed and convex, and in a Hilbert space, the intersection of any number
Rollout 15
eigenvalues of this matrix?\n\nBut eigenvalues aren't usually considered in characteristic 2 matrices unless underlying field
Rollout 13
(1 + r)^t )\n\nBut that's not particularly helpful here. Alternatively, confirm using estimates.\n\n
Rollout 11
b)(1 + c) - 2. Not sure if that helps immediately, but maybe.\n\nNow
Rollout 0
, each prime factor in 20! must go entirely to \(a\) or entirely to \(b
Rollout 8
31/126≈0.246.\n\nTo verify if that's correct.\n\n
Rollout 6
P/poly is equivalent to languages decidable by a polynomial-time Turing machine with a polynomial-length advice string
Rollout 1
use the following theorem: Any set of vectors in a Hilbert space can be isometrically embedded into
Rollout 12
/(2 sqrt(s))) ] ds\n\nHmm. Not sure if that helps. Alternatively, substitution s =
Rollout 12
(ν + u)^2 - ν2 ) = 2ν u + u2\n\n(1
Rollout 1
spaces, you can have decreasing sequences of closed convex sets with empty intersection even if finite intersections are non-empty
Top 16 Negative Activations
Rollout 11
y )\n\nBut now this seems really complicated. Perhaps there's another substitution?\n\nAlternatively, setting variables in terms
Rollout 11
) ) ]\n\nBut this is very complicated, perhaps there's no immediate simplification. However, if I
Rollout 2
impossible because the numbers are huge. So, perhaps there's a pattern or periodic behavior we can exploit when
Rollout 5
/100. \n\nFirst, let me recall some trapezoid properties. In a tr
Rollout 1
radius d / sqrt(2} approx 0.707d. So sum of radii
Rollout 6
n + c, we can invert to n �� 2^{m - c}. The number of
Rollout 1
we need to find t and C such that all points in S satisfy that equation.\n\nAlternatively, writing this
Rollout 11
2 y )\n\nBut now this seems really complicated. Perhaps there's another substitution?\n\nAlternatively, setting variables in
Rollout 8
, GG RRR\n\nk = 4: RRRR G, G RRRR\n\nSo
Rollout 12
but seems not applicable here. Wait, let me check:\n\nGiven f(ν) = ��_{
Rollout 1
two distinct vectors are orthogonal.\n\nFirst, let me recall that an orthonormal system requires two things:
Rollout 4
to C_eq and V squared. Therefore, exponents are 1 for C_eq and 2 for
Rollout 11
conjugates. Alternatively, suppose three complex numbers with arguments ��, ��, -2θ,
Rollout 2
1000. Hmm, okay, let's unpack this step by step.\n\nFirst, the problem
Rollout 10
challenging to manually spot a pattern here, but maybe there's a different approach.\n\nThe problem statement says that
Rollout 7
12 +54N/4.\n\nConvert to same denominator:\n\n62N/12 +

Layer 52

GATE_PROJ

Top 16 Positive Activations
Rollout 12
x^2)}}\] \n(where the positive value of the square root is taken) is monotonically
Rollout 12
^2)}}\] \n(where the positive value of the square root is taken) is monotonically decreasing
Rollout 12
^2x^2)}}\] \n(where the positive value of the square root is taken) is mon
Rollout 2
999 \times \cdots \times \underbrace{99\cdots9
Rollout 2
times \cdots \times \underbrace{99\cdots9}_{\text{9
Rollout 11
& \text{and} & $q=\dfrac ab+\dfrac bc+\dfrac
Rollout 1
x, x' \in S \), the vectors \( \frac{\sqrt{2}}{d
Rollout 8
\(k = 2\): \(2 \times (5 \times 4 \times 5
Rollout 2
9 \times \cdots \times \underbrace{99\cdots9}_{\text
Rollout 8
k = 2\): \(2 \times (5 \times 4 \times 5 \
Rollout 0
case \(a = b\) (which is impossible since 20! is not a square).\n\n4
Rollout 12
nu^2x^2)}}\] \n(where the positive value of the square root is taken) is
Rollout 2
999 \times \cdots \times \underbrace{99\cdots9
Rollout 1
\in S \), the vectors \( \frac{\sqrt{2}}{d}(x - y
Rollout 0
airs \(a < b\)**:\n - Since \(a\) and \(b\) are copr
Rollout 0
= b\) (which is impossible since 20! is not a square).\n\n4. **Result
Top 16 Negative Activations
Rollout 9
I'm confused. Wait squares have four sides. Let me actually assign coordinates to the square to make this
Rollout 9
. \n\nWait, in the square AIME, let me confirm the positions. If the square is labeled
Rollout 12
the dependence on ν is more explicit, perhaps.\n\nLet me see. Let's try substitution.\n\nThe integral
Rollout 11
\n\nSimilarly, simplify: + xy + terms. Let me compute each term:\n\nFirst, p = x
Rollout 3
the formula applies. So, with that said, let me proceed. Let me check again.\n\nIf R
Rollout 3
around 170 m/s. Alternatively, let's check some resource. Wait, can't access
Rollout 11
real. Perhaps we can use some relationships here. Let's think about using the fact that if z is
Rollout 2
≥3.\n\nSo similar to modulo8 analysis? Let's be precise.\n\nbreak down the terms in the
Rollout 3
is a heavy gas, so maybe at room temperature?\n\nBut surely, since no temperature is given, we
Rollout 3
So, with that said, let me proceed. Let me check again.\n\nIf Rn-22
Rollout 3
, perhaps there's a standard temperature to assume. Let me check online, but since I can't do
Rollout 5
, 181 is the correct answer. \n\n**Final Answer**\n\boxed{181
Rollout 12
recognized as a complete or incomplete elliptic integral. Let's check.\n\nIndeed, in the integral:\n\n∫
Rollout 13
to annual is $187.12.\n\n**Final Answer**\n\boxed{187
Rollout 11
2, b^2, c^2? Let's multiply numerator and denominator by appropriate terms. For
Rollout 8
we can calculate the number of such happy sequences. Let's consider two cases:\n\nCase 1: All

UP_PROJ

Top 16 Positive Activations
Rollout 0
2,3,5,7 are even?\n\nWait floor operations:\n\nFor prime 2:
Rollout 9
5)/(10 - (10 - h)) = 5/h.\n\nTherefore, equation is y
Rollout 0
(even), 2 (7) (even?), no 7 has exponent 2? Wait for
Rollout 10
7,20,22,25?\n\nWait, but n=25 is a losing
Rollout 9
)/(10 - (10 - h)) = 5/h.\n\nTherefore, equation is y -
Rollout 1
2 - 2c + d2)/2 = 0 ⇒ 0 = 0.\n\nThus
Rollout 4
14.8058)/15 = 0.39/15 ��
Rollout 9
)/(10 - (10 - h)) = (-5)/h.\n\nThe equation is y -
Rollout 0
(even?), no 7 has exponent 2? Wait for 7: floor(20/
Rollout 0
4,7:2, which are all even? Wait, 18,8,4,
Rollout 5
30625)/2] = sqrt(36250/2) =
Rollout 5
2500 = 125y\n\nSimplify:\n\ny^2 - 25
Rollout 5
125x + 9375\n\nSimplify:\n\nx^2 + 12
Rollout 9
5)/(10 - (10 - h)) = (-5)/h.\n\nThe equation is y
Rollout 7
213N -209N)/3=4N/3.\n\nTherefore:\n\n16
Rollout 5
25 + 30625)/2] = sqrt(36250/
Top 16 Negative Activations
Rollout 14
terms of $J$ and $K$.\n\nMaybe Cauchy-Schwarz on $I$
Rollout 11
b, c. Since abc = 1, maybe substitutions could help here. For example, if abc
Rollout 11
let me note that since abc = 1, maybe we can write 1/a = bc,
Rollout 2
seems impossible because the numbers are huge. So, perhaps there's a pattern or periodic behavior we can exploit
Rollout 14
is $8$ times $J K$. So, it's perhaps getting an upper bound on $I
Rollout 1
x'||2 - d2)/2.\n\nBut perhaps another way is expressing this by defining various terms.
Rollout 10
the game each move is subtract a square, so perhaps using octal games concept or mathematical induction.\n\nAlternatively
Rollout 11
is symmetric in a, b, c. So perhaps we can relate p and q through symmetric expressions.\n\n
Rollout 1
infinite set where each pair is separated by d, perhaps S must lie on such a quadratic manifold. But
Rollout 11
. Maybe that substitution isn't helpful. Instead, perhaps keeping variables as they are and using symmetries
Rollout 14
the $J$ and $K$ terms. Perhaps split the integral of $f(x)$ into two
Rollout 1
that this holds for all x in S.\n\nBut maybe we can see this as a functional equation. Let
Rollout 1
't necessarily contain a linear subspace.\n\nAlternatively, perhaps we can impose some structure. Let me fix k
Rollout 11
arguments are not 0 or π. \n\nBut maybe writing in terms of reciprocals: Since abc
Rollout 14
a product of three integrals. Wait, maybe if I represent the left-hand side as a product of
Rollout 1
how can we construct such a y?\n\nAlternatively, perhaps by considering properties of such sets S. If S

DOWN_PROJ

Top 16 Positive Activations
Rollout 10
larger square numbers alter the periodicity. Let me recast the previous positions and see the exact classifications.\n\n
Rollout 2
91 mod125: 125*7=875. 89
Rollout 2
91mod125. 125*7=875, 89
Rollout 10
0→Grundy[20}=0\n\nContinuing:\n\nn=21: subtract1→
Rollout 10
L. Then:\n\nn=34:L\n\nContinuing this, there are more L positions, but
Rollout 12
1 - k^2)}\n\]\n\nand the derivative \( \frac{dk}{d\nu}
Rollout 9
Wait, perhaps my assumption is incorrect. Let's recast this. I need to check whether when h
Rollout 14
(y) dy.\n\nIf I set c and d to normalize something.\n\nAlternatively, proceed numerically.\n\nGiven
Rollout 3
with precise calculation using more decimal places.\n\nLet me recalculate the exact value.\n\n8 * 8.
Rollout 4
025:\n\nFirst, 5025 * 1200 = 6,
Rollout 0
3, 17, 19 each have exponent 1 in 20!.\n\nSo ex
Rollout 14
chwarz comes to mind for the squared integral, but here we have three powers on the left and combined
Rollout 11
rocals. Similarly, if variables are inverses up to some rotation. \n\nHmm, this approach is
Rollout 0
product a*b=20! means that each prime's exponent in 20! is divided between
Rollout 4
02 *10, etc. Let's recalculate the derivatives values more precisely:\n\nStarting with Δ
Rollout 0
which it is not (since exponents in its prime factorization are all even? Wait, 2
Top 16 Negative Activations
Rollout 3
(225,298)= approx 474.6 m/s. So about
Rollout 3
Then sqrt(225,298)= approx 474.6 m/s.
Rollout 3
28,408.12)= 168.5\n\nThus, answer is
Rollout 3
(28408.12)= 168.546... Hence,
Rollout 3
Let's see.\n\n168^2 = (170 - 2)^2 =
Rollout 3
225,753 ) �� 475 m/s. Which is correct.\n\n
Rollout 3
0\n\nsqrt(27,950)= approximately 167.2 m/s.
Rollout 3
,408.12) �� 168.5 m/s\n\nSo approximately
Rollout 3
0; sqrt(27,950)=167.2 m/s. So around
Rollout 5
equation 1: h = (125H)/(75 + x). Plug into equation
Rollout 4
.5%), and 0.02V (0.04%), so maybe a couple of
Rollout 3
sqrt(27,950)= approximately 167.2 m/s. So ~
Rollout 3
(28,433) �� 168.6 m/s.\n\nSo my
Rollout 5
the first equation: h = (125H)/(75 + x)\nFrom the second equation
Rollout 14
)*(3^{2/3})=2^{-2/3}*3^{2/3}=
Rollout 3
68.5)=15.87/337≈0.047.

Layer 53

GATE_PROJ

Top 16 Positive Activations
Rollout 8
*9. 31 and 126 are coprime, so yeah, simplified.\n\n
Rollout 9
\n\nSo 20y = 100 => y = 5.\n\nTherefore, the y
Rollout 11
/(ab)) + (1/(ab))/a. Simplify each term: a/b is the same
Rollout 8
30240.\n\nSimplify this fraction.\n\nFirst, divide numerator and denominator by 24
Rollout 9
But the problem says this area is 80. Therefore:\n\n100 -500/h
Rollout 8
30240. But let me verify that. Yes, 10 choices for first
Rollout 8
.\n\nTherefore, 31/126. Then check if they can be simplified. Since
Rollout 11
real, then its conjugate is equal to itself. Therefore:\n\n\overline{p} = \
Rollout 7
2N)/3, R = N/3.\n\nSubstitute P and R:\n\n75*( (
Rollout 8
26. Then check if they can be simplified. Since 31 is prime. 12
Rollout 8
10P5=30240. Then 7440/302
Rollout 5
\n\nTherefore, 181 is the correct answer. \n\n**Final Answer**\n\boxed{1
Rollout 8
440 / 30240.\n\nSimplify this fraction.\n\nFirst, divide numerator and
Rollout 14
} (c - d x)^2 dx. Let me compute this. Expand (c - dx)^
Rollout 9
square, so the triangle is completely within the square. Wait, because the square spans from x=0
Rollout 2
ime, m must be ��0 mod8. Therefore, m=8n for some integer n
Top 16 Negative Activations
Rollout 13
pay for her car. The annual interest rate on the loan is $12\%$.
Rollout 5
of the legs divides the trapezoid into two regions whose areas are in the ratio $2:
Rollout 13
609^2 \approx 1.12550881, \quad
Rollout 14
2.289)^2≈5.24. Then divide by 4≈1.
Rollout 8
and $5$ green cards. She shuffles the $10$ cards and lays out $5
Rollout 13
� 6000 * 1.57351936 ��
Rollout 13
She makes no payments for 4 years, but has to pay back all the money she owes at the
Rollout 8
cards and $5$ green cards. She shuffles the $10$ cards and lays out $
Rollout 3
921;162^2=26,244. So it's between
Rollout 13
6000 * 1.57351936 �� 9
Rollout 6
bb{N} \mapsto \mathbb{N}$ such that for each $n
Rollout 13
4 years. How much more money will she owe if the interest compounds quarterly than if the interest compounds annually
Rollout 13
0609^2 \approx 1.12550881, \
Rollout 3
12 * 270 = 17,958.24, plus
Rollout 3
4. 169^2 = 28,561. So sqrt(2
Rollout 8
but RRRGR will not. The probability that Kathy will be happy is $\frac{m}{n

UP_PROJ

Top 16 Positive Activations
Rollout 4
0.003605^2 = approx 0.000013
Rollout 12
(1 - t2)\n\nSo denominator becomes sqrt( ( (t2 - ν2)/ν2
Rollout 4
6055555)^2= 0.00001300
Rollout 3
,820.576 �� 0.697433 ��
Rollout 5
2:\n\n(x + 175)(H - (125H)/(75 + x
Rollout 6
one and keep the combined set sparse? Let's think. Suppose each S_i has at most p(n
Rollout 4
_eq is ΔC_eq / C_eq = sqrt( [ (C2 / (C1 (C
Rollout 5
2:\n\n(x + 175)(H - (125H)/(75 +
Rollout 5
So:\n\n(x + 175)(H - (125H)/(75 + x
Rollout 14
/2} f(x)^2 dx. Not sure.\n\nAlternatively, think about splitting f(x) as
Rollout 3
kg/mol.\n\nNow, the average speed is sqrt(8RT/(πM)).\n\nBut if the
Rollout 3
.512\n\nThen, 66.512 * 298:\n\nLet
Rollout 4
36055555)^2= 0.0000130
Rollout 13
this:\n1.12*1.12: 1*1 = 1,
Rollout 3
28 g/mol, the average speed is sqrt(8*8.314*29
Rollout 3
.0\n\n168.5^2=168^2 + 2*1
Top 16 Negative Activations
Rollout 11
Answer**\n\boxed{(-1, 3)}\n\n<|im_start|>answer\nGiven three complex numbers \(a,
Rollout 13
**\n\boxed{187.12}\n\n<|im_start|>answer\nTo determine how much more money Jo
Rollout 2
**Final Answer**\n\boxed{109}\n\n<|im_start|>answer\nTo find the remainder when \(9
Rollout 9
.\n\n**Final Answer**\n\boxed{25}\n\n<|im_start|>answer\nSquare \(AIME\) has sides
Rollout 5
**Final Answer**\n\boxed{181}\n\n<|im_start|>answer\nLet the shorter base be \( b
Rollout 7
boxed{24}, \boxed{36}\n\n(b) No solution exists.\n\n<|im_start|>answer\nGiven
Rollout 8
**Final Answer**\n\boxed{157}\n\n<|im_start|>answer\nKathy has 5 red cards
Rollout 10
player can win in these cases.\n\n**Final Answer**\n\boxed{There are infinitely many initial situations where
Rollout 0
**Final Answer**\n\boxed{128}\n\n<|im_start|>answer\nGiven a rational number between 0
Rollout 10
situations where the second player can force a win.}\n\n<|im_start|>answer\nTo prove that there are infinitely many
Rollout 15
guarantee that all lamps will be off after some time?\n<|im_start|>assistant\n<|im_start|>think\nAlright, let's
Rollout 8
2: RRGGG, GGGGRR\n\nWait, wait, maybe I need to reorgan
Rollout 8
= 157.\n\n**Final Answer**\n\boxed{157}\n\n<|im_start|>answer\nK
Rollout 15
\) for some non-negative integer \( k \).\n\n**But Wait:**\nAfter reconsidering from some
Rollout 10
second player can win no matter how his opponent plays.\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I
Rollout 14
) \, dx \right) }\n\]<|im_end|>

DOWN_PROJ

Top 16 Positive Activations
Rollout 2
109 mod125.\n\nSo summary so far:\n\nP≡5 mod8\n\nand P
Rollout 2
=3 onward are -1 mod8. So starting from k=3 up to k=99
Rollout 7
-54) = 2/3\n\nMultiply both sides by 12:\n\nA -5
Rollout 4
{0.88\%}.\n\n<|im_start|>answer\nTo find the percentage error in the calculation of
Rollout 3
.028) ) �� sqrt( (19872.7 ) / (
Rollout 1
⟩ = k for all x ∈ S.\n\nThis is a system of equations. For all x in S
Rollout 13
to annual is $187.12.\n\n**Final Answer**\n\boxed{187
Rollout 7
74 -A)*(N/12)\n\nMultiply both sides by12:\n\n-24N
Rollout 5
+ x)) = 125H\n\nMultiply out:\n\n(x + 175)H
Rollout 2
891*(1000 -1)=891*1000 -8
Rollout 2
891*(1000 -1)=891,000 -89
Rollout 3
0.222)) �� sqrt( (8*8.314*29
Rollout 12
on } (0, 1)}.\n\n<|im_start|>answer\nTo prove that the function \( f(\nu
Rollout 5
100$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. I need to solve
Rollout 11
\boxed{(-1, 3)}\n\n<|im_start|>answer\nGiven three complex numbers \(a, b,
Rollout 13
the answer is $187.12.\n\nJust to ensure there were no calculation errors, let
Top 16 Negative Activations
Rollout 3
. At STP, the molar volume is 22.4 liters, which is for ideal
Rollout 3
8,436). Let's see, 168^2 = 28,
Rollout 3
applies. So, with that said, let me proceed. Let me check again.\n\nIf Rn-
Rollout 2
is the case?\n\nWait, but this conflicts with our result for modulo8. If P≡5 mod
Rollout 2
0^{k-3}≡0*10^{k-3}=0 mod100
Rollout 0
,3,5,7 are even?\n\nWait floor operations:\n\nFor prime 2: 20
Rollout 15
ON today), irrespective of multiple counts.\n\nWait a careful analysis.\n\nHere's how we can bridge:\n\nThe
Rollout 0
9 have exponent 1.\n\nSo primes 2: 18, 3: 8,
Rollout 3
,408.12} \approx 168.5 \, \text{
Rollout 14
3})=2^{-2/3}*3^{2/3}= (3^{2}/2
Rollout 7
the original scores. But this is ambiguous.\n\nAn alternative interpretation is that when they increased all scores by5
Rollout 6
Because in the problem statement, given S_1,...,S_k (finite k), we need to
Rollout 12
+ b)} } \n\nBut maybe that's diverging.\n\nAlternatively, with substitution t = sinθ,
Rollout 3
original context, the temperature is standard. Maybe just proceed with 298 K and note the assumption
Rollout 3
222 kg/mol. Correct.\n\nUse R=8.314 J/(mol·K
Rollout 11
1, their inverse conjugates sum is real. Not sure.\n\nGiven the problem constraints and difficulty, perhaps

Layer 54

GATE_PROJ

Top 16 Positive Activations
Rollout 4
0.02 / 5.00 = 0.004.\n\nNow, since
Rollout 4
327 / 1200 �� 0.003606.
Rollout 8
31 +126=157.\n\nBut I want to cross-verify.\n\nWait
Rollout 13
to annual is $187.12.\n\n**Final Answer**\n\boxed{187
Rollout 4
the percentage error is approximately 0.88%.\n\n**Final Answer**\nThe percentage error in the calculation
Rollout 4
.327 / 1200 = let's compute that more accurately.\n\n4.3
Rollout 13
the answer is $187.12.\n\nJust to ensure there were no calculation errors, let
Rollout 4
.003606. So, approximately 0.36%.\n\nNow, moving on
Rollout 4
.327 / 1200 �� 0.003606
Rollout 4
= 10 / 2000 = 0.005. Similarly, for
Rollout 5
5 is 181. Therefore, final answer is 181. \n\nI think that
Rollout 4
be 0.195 /15 =0.013, 1.3
Rollout 8
the probability is 31/126, so m=31, n=12
Rollout 13
144 = 1.2544. But wait, actually that method is confusing.
Rollout 5
, 181 is the correct answer. \n\n**Final Answer**\n\boxed{181
Rollout 4
0.008774, or 0.8774%\n\nThus,
Top 16 Negative Activations
Rollout 14
8, but our computed maximum was only ~2.25. Thus the inequality is actually not tight
Rollout 6
N}}$ with $\alpha_n \in \{0,1\}^{p(n
Rollout 3
and pressure, with boiling point at -61.7 °C, so at STP (0
Rollout 14
0^\infty f(x)^2 dx \right) \left(\int_0^\infty
Rollout 3
pi*0.028= ~0.08796\n\n19,8
Rollout 13
05229 + 0.03582157 �� 1
Rollout 13
27407 + 0.03477822 �� 1
Rollout 13
3371188 + 0.0440560113
Rollout 14
�1.31 which cubes to ~2.25, which is way small than8,
Rollout 12
}{\nu}} \frac{dx}{\sqrt{(x^2-1)(1-\nu
Rollout 2
Find the remainder when $9 \times 99 \times 999 \times \cd
Rollout 13
1637789 + 0.0403174913
Rollout 5
two regions whose areas are in the ratio $2: 3$ . Let $x$ be the
Rollout 13
3386923 + 0.0415270160
Rollout 8
<|im_start|>user\nKathy has $5$ red cards and $5$ green cards. She shuffles
Rollout 13
86923 + 0.04152701608≈

UP_PROJ

Top 16 Positive Activations
Rollout 5
)) = 125H\n\nLet me factor H out:\n\n(x + 175)
Rollout 7
18N =56N/3\n\nFactor x:\n\nx(A -54) +1
Rollout 5
5 + x) = 125\n\nFactor the second term:\n\nLeft side = (x +
Rollout 11
r3 +1/r3 )=0\n\nFactor out 2i sinθ:\n\n2i sin
Rollout 11
2) sinθ cosθ = 0\n\nFactor out sinθ:\n\nsinθ [3 + (
Rollout 5
75 + x)\n\nExpand the left-hand side:\n\nx^2 - 50x +
Rollout 7
4) = 2N/3\n\nDivide both sides by N (assuming N ≠0,
Rollout 9
^2 + (10 - y)^2]\n\nSquaring both sides:\n\nh^2 +
Rollout 5
(75 + x)\n\nExpand the left-hand side:\n\nx^2 - 50x +
Rollout 11
i sin2θ) = 0\n\nDivide by 2i:\n\n3 sinθ + (
Rollout 7
66P + 66R\n\nSubtracting 66P and 56R
Rollout 7
6(P + R)\n\nLet me simplify this equation:\n\n71P + 56R =
Rollout 5
0 = 125y\n\nBring all terms to left side:\n\ny^2 - 2
Rollout 7
= 66(P + R)\n\nLet me simplify this equation:\n\n71P + 56
Rollout 5
00) = 125y\n\nFactor left-hand side:\n\n(y + 100
Rollout 7
R = 66P + 66R\n\nSubtracting 66P and
Top 16 Negative Activations
Rollout 8
m = 31\) and \(n = 126\). Thus, \(m +
Rollout 0
6}{2} = 128\).\n\nThus, the number of rational numbers between 0
Rollout 9
h}x + 10 - \frac{50}{h}\).\n\n5. **Intersection
Rollout 1
formed by translating and scaling the set \( S \).\n\n\[\n\boxed{\text{Such a point
Rollout 13
= 187.12\n\]\n\n\[\n\boxed{187.1
Rollout 11
\(p\) and \(q\) are real.\n\n1. **Key Observations**:\n - Since
Rollout 9
EM\) in \(\triangle GEM\).\n\n1. **Vertices of the Square**:\n -
Rollout 9
implies h = 25\n \]\n\nThus, the length of the altitude to \(EM
Rollout 8
, \(m + n = 31 + 126 = 157\).\n\n
Rollout 9
\frac{500}{h} = 80 \implies \frac{50
Rollout 0
, the number of rational numbers between 0 and 1 for which the product of the numerator and denominator
Rollout 2
\mod 1000\n \]\n\nThus, the remainder when the product is divided by
Rollout 8
126 = 157\).\n\n\[\n\boxed{157}\n\
Rollout 0
a \times b = 20!\).\n\n1. **Prime Factorization of 20!
Rollout 15
ly turns off.\n - For \( n = 2 \) and higher, testing specific configurations shows
Rollout 15
\[\n\boxed{2^k}\n\]\n\nfor non-negative integers \( k \).<|im_end|>

DOWN_PROJ

Top 16 Positive Activations
Rollout 3
� 28,436. Then take square root.\n\nsqrt(28,43
Rollout 11
, then A + B + C = 2πik for some integer k. But since a,
Rollout 11
-real c as well.\n\nAlternatively, take a = re^{iθ}, b = re^{iθ
Rollout 4
.00007696\n\nTake the square root: sqrt(0.00
Rollout 3
,408.12\n\nNow, take the square root of 28,40
Rollout 3
4, as 161^2=25,921;162^
Rollout 3
921;162^2=26,244. So it's between
Rollout 14
}. Then 12^{2/3}=2^{4/3} *3^{2/
Rollout 13
annual is $187.12.\n\n**Final Answer**\n\boxed{187.
Rollout 14
/3}/4 )^3 = (12^{2/3})^3 /4^
Rollout 4
percentage error is approximately 0.88%.\n\n**Final Answer**\nThe percentage error in the calculation of
Rollout 11
take a, b, c such that a = e^{iθ}, b = e^{iθ
Rollout 11
} ] =0\n\nUsing Euler’s formula, e^{iΦ} - e^{-iΦ}
Rollout 4
2 = 0.00001296\n\nAdding those together: 0.
Rollout 4
4)^2 )=sqrt(0.1296 +0.64)=sqrt(
Rollout 4
max-min approach assumes that all errors contribute in the same direction, which is more of a worst-case scenario
Top 16 Negative Activations
Rollout 5
base. The segment that joins the midpoints of the legs divides the trapezoid into two regions
Rollout 10
squares, etc. Such a set {ni} has the property that no two differ by a square,
Rollout 1
we have a set of vectors with equal norms and their differences have equal norms, this often relates to them
Rollout 0
the problem reduces to finding the number of coprime pairs \((a, b)\) where \(
Rollout 1
use that in H, you can have orthonormal systems of arbitrary cardinality.\n\nSo, using Z
Rollout 1
Hilbert space, one can have an orthonormal basis of arbitrary cardinality or an appropriate system,
Rollout 1
in non-separable Hilbert spaces, you have uncountable orthonormal bases, while in separ
Rollout 1
need the family to be "centered" (has the finite intersection property) and maybe in a weak
Rollout 0
both.\n\nTherefore, the number of such coprime pairs is \(2^k\), where \(
Rollout 0
20!. So they form a coprime pair whose product is 20!. Such pairs
Rollout 1
caled. Thus, this becomes, an orthonormal family. So in infinite dimensions, such a family
Rollout 1
certain conditions hold, such as the family of spheres has the finite intersection property.\n\nBut the problem says H
Rollout 1
but in a Hilbert space, any orthonormal system must be countable because the space is separ
Rollout 4
62 +0.0082)=sqrt(0.0000129
Rollout 15
Alternatively, perhaps answer is that n must be a power of2? BUT n=1 is OK
Rollout 1
" or "orthonormal basis" even if uncountable. However, typically in non-separable

Layer 55

GATE_PROJ

Top 16 Positive Activations
Rollout 4
is approximately 0.88%.\n\n**Final Answer**\nThe percentage error in the calculation of the energy
Rollout 4
[\n\boxed{0.88\%}\n\]<|im_end|>
Rollout 3
69} \, \text{m/s}\n\]<|im_end|>
Rollout 15
of lamps \( n \) is a power of 2. This result is derived from properties of linear
Rollout 3
, \text{m/s}\).\n\n**Final Answer:**\n\[\n\boxed{169
Rollout 8
m = 31\) and \(n = 126\). Thus, \(m +
Rollout 4
) using error propagation. The partial derivatives are:\n\n\[\n\frac{\partial C_{\text{
Rollout 9
h}x + 10 - \frac{50}{h}\).\n\n5. **Intersection
Rollout 15
ly turns off.\n - For \( n = 2 \) and higher, testing specific configurations shows
Rollout 1
such spheres is non-empty.\n\n5. **Conclusion**:\n - The existence of such a point \(
Rollout 4
_{\text{eq}} \) is:\n\n\[\n\frac{\Delta C_{\text{
Rollout 14
with appropriately chosen vectors / weighet\n\n**Final Answer**\n\boxed{ \left(\int_0
Rollout 12
otonically decreasing on } (0, 1)}\n\]<|im_end|>
Rollout 3
\text{m/s}\).\n\n**Final Answer:**\n\[\n\boxed{169}
Rollout 13
550881, \quad \text{and recursively calculating up to } 1.
Rollout 8
of the same color).\n - Total for Case 1: \(2 \times 5! =
Top 16 Negative Activations
Rollout 12
(t2 - ν2) becomes (ν2 - ν2) =0, so that first
Rollout 12
( (1/ν) sqrt( (t2 - ν2)(1 - t2) ))
Rollout 12
) / sqrt( ( (t/ν)^2 -1 )(1 - t^2) )\n\n
Rollout 9
ating GE distance:\n\nGE is from G (10 - h, y) to E (10
Rollout 12
ν^1 [1 / sqrt( (t2 - ν2)(1 - t2) )
Rollout 12
to 1 of dt / sqrt( (t2 - ν2)(1 - t2) )\n\n
Rollout 12
, at t=ν, the term (t2 - ν2) becomes (ν2 - ν
Rollout 12
1 / sqrt( ( (ν + u)^2 - ν2 ) (1 - (ν +
Rollout 9
h, and the location of G is (10 - h, y). But the triangle being is
Rollout 12
t=ν, it's formally ( (ν2 - ν2) * (1 - ν2
Rollout 12
}^1 1 / sqrt( (t2 - ν2)(1 - t2) )
Rollout 12
1}{\sqrt{( (1/\nu)^2 - 1)(1 - \nu^2
Rollout 12
2)\n\nSo denominator becomes sqrt( ( (t2 - ν2)/ν2 ) (1 -
Rollout 14
, through substitution or integration by parts.\n\nWait, another idea is to use the Cauchy-Sch
Rollout 12
)K(k)~ x*(- (1/2) lnx ) ~ - (x lnx
Rollout 7
(but <65) and (N/3 -x) students with original scores <60

UP_PROJ

Top 16 Positive Activations
Rollout 2
Modulo 125 Calculation:**\n - Each term \(10^k - 1
Rollout 2
is \(2^k - 1\).\n - For \(k \geq 3\
Rollout 15
neighbors is on at time \( t \).\n - This rule translates to the next state being the sum
Rollout 9
G\) to \(EM\) be \(h\). The coordinates of \(G\) are \((1
Rollout 15
1. **Modeling the System**:\n - For each lamp \( i \), the next state
Rollout 12
\( 0 < k < 1 \). Therefore, the numerator in the expression for \( f'
Rollout 4
.16\n\]\n\nThe uncertainties are:\n\n\[\n\Delta C_{\text{eq}}
Rollout 1
. **Orthonormal Conditions**:\n - For each \( x \in S \), the vector
Rollout 7
a)\n\n1. **Original equations:**\n - Total score: \( 66N =
Rollout 6
\) with the binary representation of \( i \). Since each \( S_i \) is sparse,
Rollout 5
\implies b = 75\n\]\nThus, the bases are 75 and
Rollout 10
-positions)**:\n - A position is losing if all possible moves lead to winning positions.\n -
Rollout 6
). Since each \( S_i \) is sparse, for each length \( n \), the number of
Rollout 1
1. **Orthonormal Conditions**:\n - For each \( x \in S \), the
Rollout 15
sum modulo 2 of the neighbors' current states. For end lamps, the next state is simply the
Rollout 14
= I_1 + I_2.\n\]\n\nUsing the Cauchy-Schwarz inequality for
Top 16 Negative Activations
Rollout 8
the number of ordered sequences is 10P5. If they are indistinct, then the
Rollout 8
5 cards in order would be 10P5, as earlier.\n\nBut the problem here is that
Rollout 8
0 different cards, hence the number of ordered sequences is 10P5. If they are ind
Rollout 8
7 * 6 = 30240. However, in the color sequence RRRRR
Rollout 0
assigned to a or b, so 2^3=8 coprime pairs. Then half of
Rollout 8
10P5 = 30240. However, each color sequence (like RRGG
Rollout 7
Thus N=36 works. So possible N are 12,24,36.
Rollout 0
be 2^8 /2 =128. So 128 coprime pairs
Rollout 8
number of possible sequences is 30240.\n\nNow, we need to count how many of
Rollout 8
to our method:\n\nTotal sequences: 4P2=12\n\nHappy sequences: All red (
Rollout 8
replacement and the order matters. Alternatively, the number is 10P5 = 10!
Rollout 0
ime assignments (with order) is 2^3=8, but how many of those assignments result
Rollout 0
pairs (a,b) would be 2^8 = 256. However, because
Rollout 8
where she lays out 5 cards in order would be 10P5, as earlier.\n\nBut
Rollout 5
1. Therefore, the answer is 181. \n\nWait, but let me verify once again
Rollout 2
the remainder upon division by 1000 is109. But wait, but here's

DOWN_PROJ

Top 16 Positive Activations
Rollout 7
42)=14N/3\n\nMultiply both sides:\n\n(A -42)= (14
Rollout 11
sum of the variables plus the sum of their products two at a time. That form makes me think of
Rollout 8
5 red cards, but since they are identical? Wait, wait, no. Wait, the problem says
Rollout 10
25: moves can take 1,4,9,16,25.\n\n25
Rollout 10
=3、2+4=6、2+9=11, etc. The earliest a
Rollout 1
other intersection theorem, which states that if we have a family of closed convex sets where every finite intersection is
Rollout 5
+ x) = 125\n\nMultiply both sides by (75 + x):\n\n(x +
Rollout 2
9 mod1000.\n\nWait, but hold on, earlier we got x=125
Rollout 10
4 is winning.\n\nn=15: subtract 1=14 (win), subtract 4
Rollout 10
≡1 mod5, also winning.\n\n20-9=11≡1 mod5, winning
Rollout 7
) = 2N/3\n\nDivide both sides by N (assuming N ≠0, which
Rollout 10
9 is winning.\n\nn=10: subtract 1=9 (winning), subtract 4
Rollout 6
the input i is between 1 and k.\n\nWait, but how does S know about k? Because
Rollout 7
42*(N/12)\n\nDivide both sides byN:\n\nA/12=1
Rollout 5
)] = 125H\n\nDivide both sides by H (assuming H ≠ 0):\n\n
Rollout 6
3 would give p(n) = 4 and 9, different. So if p is injective
Top 16 Negative Activations
Rollout 8
4/3024=744÷24=31; 302
Rollout 15
one neighbor is on, then next state is on. So, for the lamp i, next state is
Rollout 8
= Let's check by dividing numerator and denominator by 240: 7440÷
Rollout 1
/ 2,\n\nwhich is ⟨x, x⟩ - 2⟨x, y⟩ + ⟨
Rollout 3
textbooks usually use 298 K (unless stated otherwise), perhaps answer around \boxed{16
Rollout 14
^3 ≤8, which is equivalent to I≤2. So how to prove that I ≤2
Rollout 15
of path graphs. Online resources indicate that for the linear CA where next state is the sum mod2 of
Rollout 15
so, then n is acceptable (all lamps die), otherwise, it's not.\n\nThis thus connects the
Rollout 3
is the molar mass. Then there's the average speed, which is sqrt(8RT/(π
Rollout 6
. If yes, then x was in S_i, otherwise not. But is the encoding invertible?
Rollout 6
) consists of all tuples \((i, x)\) where \( x \in S_i \)
Rollout 1
Assume H has an orthonormal basis. If we have such a y, then the vectors { sqrt
Rollout 1
Hilbert space, given two vectors u and v, the parallelogram law says that ||u +
Rollout 8
all red or all green sequences are also acceptable, like RRRRR or GGGGG.\n\nSo in
Rollout 15
the single neighbor is on, next state is on; else, off. Therefore, end lamp's next
Rollout 4
597*24=14,328 and 597*0.

Layer 56

GATE_PROJ

Top 16 Positive Activations
Rollout 13
to annual is $187.12.\n\n**Final Answer**\n\boxed{187
Rollout 10
=1: First player takes 1 pebble (since 1 is a square), leaving 0
Rollout 0
=2^{8-1}=128.\n\n**Final Answer**\n\boxed{128
Rollout 2
09.\n\nTherefore, answer:109.\n\nBut let me recheck all steps once again to
Rollout 8
m=31, n=126.\n\nWait, but let's confirm. Wait, when
Rollout 8
1 +126= 157.\n\n**Final Answer**\n\boxed{157
Rollout 2
=1000n +109, meaning x≡109 mod100
Rollout 2
product is divisible by 1000, so the remainder is 0. However, given that
Rollout 11
)=(-1,3). Hence, ordered pair is \boxed{(-1, 3)}.\n\n
Rollout 8
31, n=126, which are coprime, so sum m +n=
Rollout 8
126 are coprime, so yeah, simplified.\n\nTherefore, the probability is 31
Rollout 4
the percentage error is approximately 0.88%.\n\n**Final Answer**\nThe percentage error in the calculation
Rollout 15
off.\n\nt=2: O F F F\n\nThis is same as t=0. Thus,
Rollout 13
the answer is $187.12.\n\nJust to ensure there were no calculation errors, let
Rollout 2
same as k=3, -1 mod8\n\nAll terms from k=3 onward are -1
Rollout 5
, 181 is the correct answer. \n\n**Final Answer**\n\boxed{181
Top 16 Negative Activations
Rollout 14
I to F and K.\n\nNo, that seems not immediately helpful. \n\nWait, have you consider
Rollout 1
2)} / 2.\n\nBut perhaps it's hard.\n\nWait, back to the equation:\n\nFor y
Rollout 1
orthocentric system or something similar?\n\nLet me think.\n\nSuppose we fix y. Then the conditions
Rollout 14
say f(x) = g(y ). But likely not helpful.\n\nEarlier tried homogeneous functions already.\n\nAlternatively,
Rollout 12
the Jacobi elliptic function. Wait, maybe not necessary. Alternatively, use a trigonometric substitution
Rollout 11
a/c = |b|^2. But not sure.\n\nSimilarly, set each term in q equal to
Rollout 14
a^{-2}) K]^ some combination... probably not.\n\nAlternatively, use Young's inequality which relates products
Rollout 1
a_i such that y exists.\n\nBut seems difficult to see.\n\nAlternative idea. From the properties we earlier
Rollout 9
to EM, h can be anything. Let's think. If h is less than 10,
Rollout 14
and 3/2.\n\nWait, let me think about the three-way Hölder: for three functions
Rollout 10
greater than k2 for k >2. Might not help.\n\nAnother angle: If n is such that
Rollout 14
functions where one involves $x$?\n\nLet me think. Let's consider $f(x) = f
Rollout 1
coefficients a_i such that y exists.\n\nBut seems difficult to see.\n\nAlternative idea. From the properties we
Rollout 1
be uncountable. Hmm.\n\nBut let's think: Suppose that S is a set where all pairwise
Rollout 14
}$, but that feels like a stretch and may not directly lead to the terms we have.\n\nWait,
Rollout 11
now this seems really complicated. Perhaps there's another substitution?\n\nAlternatively, setting variables in terms of exponentials

UP_PROJ

Top 16 Positive Activations
Rollout 0
2,3,5,7 are even?\n\nWait floor operations:\n\nFor prime 2:
Rollout 2
999*10 +9, wait, perhaps 9999=9*1
Rollout 5
(725). Is that right?\n\nWait, let me check the calculations again step by step.
Rollout 4
5000^2) )^2 )\n\nWait, this is another way of writing the terms
Rollout 4
) * 15 )^2\n\nHmm, let's compute these step by step.\n\nFirst,
Rollout 3
416*0.222) )\n\nFirst compute numerator:\n\n8*8.31
Rollout 5
* sqrt(725). Is that right?\n\nWait, let me check the calculations again step by
Rollout 15
But this seems contradictory. Wait, maybe a mistake.\n\nWait, let's think again. Let's map
Rollout 8
2: 2400.\n\nWait, but hold on. Let me check for k=
Rollout 2
-1 for anyk is always even? No. When k=1: 9 is odd.
Rollout 2
0^k is 0 mod125?\n\nWait, 10^1 =10
Rollout 3
π * 0.222)) Hmm, okay. Let me calculate.\n\nFirst, 8
Rollout 0
2,3,5,7 are even?\n\nWait floor operations:\n\nFor prime 2: 2
Rollout 13
255 + 255^2.\n\nBut perhaps break down using standard multiplication:\n\nBut perhaps
Rollout 13
calculation gives around 1.5735?\n\nAlternatively, I could use a calculator step here,
Rollout 2
1=101*11? Wait: 1111=101*
Top 16 Negative Activations
Rollout 14
sqrt{J}}{2\sqrt{A}} - \frac{K}{A^2}
Rollout 12
K}{dk} = \frac{E(k) - (1 - k^2)K(k
Rollout 12
d/dk K(k) = [ E(k) - (1 -k2)K(k)
Rollout 1
are non-empty. So perhaps we need the family to be "centered" (has the finite intersection
Rollout 12
E(\sqrt{1 - \nu^2}) - \nu^2 K(\sqrt{1
Rollout 14
I^2 - 2 \lambda f(x) - \mu x = 0\n$$\n\n
Rollout 12
E(\sqrt{1 - \nu^2}) - \nu^2 K(\sqrt{1
Rollout 0
.\n\nAnd this number is equal to 2^{k -1}, where k is the number of distinct
Rollout 12
(nu) <0 if [ nu K(k) - E(k)/nu ] <0, since
Rollout 12
) ] K(k)\n\n= [ nu K(k) - E(k)/nu ] / (1 -
Rollout 1
in particular, the set S would be in a one-to-one correspondence with an orthonormal system.\n\n
Rollout 12
are:\n\ndK/dk = [ E(k) - (1 -k2)K(k)
Rollout 9
\right) \, dx = 100 - \frac{500}{h}\n
Rollout 12
dK}{dk} = \frac{E(k) - (1 - k^2)K
Rollout 4
C2))\n\n= dC1 [1/C1 - 1/(C1 + C2)]
Rollout 6
most p(n) strings of length n. Since there are k sparse sets, maybe we combine them into

DOWN_PROJ

Top 16 Positive Activations
Rollout 6
with a sparse oracle. (This is the Meyer, I think, theorem.) So perhaps this is the
Rollout 4
percentage error is approximately 0.88%.\n\n**Final Answer**\nThe percentage error in the calculation of
Rollout 8
+126= 157.\n\n**Final Answer**\n\boxed{157}\n\n
Rollout 0
2^{8-1}=128.\n\n**Final Answer**\n\boxed{128}\n\n
Rollout 13
annual is $187.12.\n\n**Final Answer**\n\boxed{187.
Rollout 14
warz inequality with appropriately chosen vectors / weighet\n\n**Final Answer**\n\boxed{ \left(\int
Rollout 12
1.\n\nHence, the proof is complete.\n\n**Final Answer**\n\n\boxed{f(\nu)
Rollout 3
\boxed{169} m/s.\n\n**Final Answer**\nThe mean molecular speed of radon
Rollout 5
181 is the correct answer. \n\n**Final Answer**\n\boxed{181}\n\n
Rollout 1
exists as described}}" or as per problem statement.\n\n**[Final Answer]**\n\boxed{\text{
Rollout 9
the altitude being h=25.\n\nPerfect.\n\n**Final Answer**\n\boxed{25}\n\n<|im_start|>
Rollout 15
ynomials.\n\nIn paper titled "Nilpotent\n\n**Final Answer**\nAll positive integers that are powers of
Rollout 6
polynomial-time machine \( M \). By the Meyer's theorem equivalence between \( \textbf{P}/
Rollout 6
machine with a sparse oracle. (This is the Meyer, I think, theorem.) So perhaps this is
Rollout 11
�_b + ��_c �� 0 mod 2π. So, ��_a +
Rollout 15
for n=2. Therefore, n=2 is not acceptable.\n\nHmm, but wait, is it
Top 16 Negative Activations
Rollout 11
1/(a2 z )* a\n\nSimplify term by term:\n\n- a\n- a z\n
Rollout 4
)^2\n\nHmm, let's compute these step by step.\n\nFirst, compute the coefficients for Δ
Rollout 12
) * ( - nu /k )\n\nSimplify step by step:\n\nFirst, substitute 1 -k
Rollout 4
=5000\n\nCompute b2/(a + b)^2 = (3000
Rollout 4
.87%.\n\nBut wait, let's go step by step for verification once more.\n\nFirst, Compute
Rollout 4
Calculating the first term: (b2/(a + b)^2 * Δa)^2 =
Rollout 6
length. But maybe if we fixed the encoding to use the same number of bits for each n. Wait
Rollout 4
_eq = sqrt [ ( (b2 / (a + b)^2 ) * Δa )^
Rollout 3
π*M))\n\nLet's calculate the numerator and denominator step by step.\n\n8*R*T = 8 *
Rollout 7
there must be a mistake here. Let me verify step by step.\n\nWe derived:\n\nA*x +4
Rollout 14
+ d^2 x^2. Then integrate term by term:\n\nc^2*(c/d)
Rollout 4
.6 pF\n\nSecond term a2/(a +b)^2= (2000
Rollout 4
C_eq = sqrt( ( (b2)/(a + b)^2 * Δa )2 +
Rollout 4
, but because (C1*C2)/(C1 + C2), if both C1 and C
Rollout 13
compute 1.03^8 as accumulating step by step.\n\nBut perhaps instead of using exponentiation
Rollout 11
}))(re^{iθ})\n\nLet me compute step by step:\n\n- a + b = 2

Layer 57

GATE_PROJ

Top 16 Positive Activations
Rollout 12
, and when x =1, t= ν.\n\nBut then x = t/ν, dx =
Rollout 12
- t2) ) / t ] dθ\n\nTherefore, dt / [ sqrt( (t2
Rollout 12
t2) ) ] = dθ / t\n\nSo, rearranged, the integrand dt /
Rollout 12
nu /k^2 ) / ( nu2 )\n\n= [ E(k) - nu2 K(k
Rollout 8
,k) * P(5,5 -k)]\n\nWait, for k=1:\n\nNumber for each
Rollout 12
- ν2)(1 - t2) )) )\n\nSimplify the 1/ν and ν canc
Rollout 11
(\overline{a}\overline{b})) ) = \overline{b} * \
Rollout 4
2000 * 5000) = (30,000)/(1
Rollout 5
100))/2 * (h/2) = (2b + 150)/
Rollout 12
] / [ (1 -nu2)nu )\n\nDenominator: (1 -nu2)nu
Rollout 12
2 -1 )(1 - t^2) )\n\nHmm. Let's compute the expression under the square
Rollout 9
would be inside. If G is inside the square.\n\nWait, but earlier when h=10,
Rollout 4
dC2 / (C1 + C2))\n\n= dC1 [1/C1 -
Rollout 12
* (1/(2 sqrt(s))) ] ds\n\nHmm. Not sure if that helps. Alternatively,
Rollout 0
then we need to divide the total by 2.\n\nWait, but the problem says "how many rational
Rollout 12
) ) ] dt, but we have to check:\n\nWait, from the substitution. If t is expressed
Top 16 Negative Activations
Rollout 1
$ \mathcal{H}$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to show
Rollout 3
radon (Rn) in m/s\n<|im_start|>assistant\n<|im_start|>think\nOkay, I need to
Rollout 7
of non-promoted $47$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I'm
Rollout 14
fty xf(x) dx \right) $$ \n<|im_start|>assistant\n<|im_start|>think\nOkay, so I have
Rollout 6
$L$ . [/*]\n<|im_start|>assistant\n<|im_start|>think\nOkay, let me try to work
Rollout 7
-promoted $47$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I'm trying to
Rollout 2
$1000$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I have this problem here
Rollout 3
Rn) in m/s\n<|im_start|>assistant\n<|im_start|>think\nOkay, I need to calculate the mean
Rollout 7
oted $47$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I'm trying to solve
Rollout 11
)$ .\n\n*David Altizio*\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I have this problem here
Rollout 14
(x) dx \right) $$ \n<|im_start|>assistant\n<|im_start|>think\nOkay, so I have to prove
Rollout 0
{}!$ be the resulting product?\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I need to figure
Rollout 12
$ . [P. Turan]\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to prove
Rollout 2
by $1000$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I have this problem
Rollout 9
$EM$ in $\triangle GEM$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need
Rollout 6
language $L$ . [/*]\n<|im_start|>assistant\n<|im_start|>think\nOkay, let me try

UP_PROJ

Top 16 Positive Activations
Rollout 9
to make this clearer.\n\nLet's place the square on a coordinate system. Let's say point A is
Rollout 9
on the left side of EM. Wait, in our coordinate system, EM is at x=10
Rollout 9
make this clearer.\n\nLet's place the square on a coordinate system. Let's say point A is at
Rollout 2
? mod 1000\n\nSolve this system using Chinese Remainder Theorem.\n\nWe need
Rollout 11
must have non-real coefficients, making it hard to apply that.\n\nAlternatively, think of log variables? If
Rollout 8
of possible sequences would be different. But I think in probability problems with colored cards, unless specified, we
Rollout 9
. Wait squares have four sides. Let me actually assign coordinates to the square to make this clearer.\n\nLet
Rollout 11
anywhere. Finally, think of the answer. Since in olympiad problems like this sometimes uses (p
Rollout 11
coefficients and equations. But without real roots. But from Descarte’s signs, if the polynomial with real
Rollout 0
128. But hold on, is this correct?\n\nWait, perhaps not quite. Since not
Rollout 2
the P modulo1000. But from the problem, rather than factorizing each term, might
Rollout 9
the base of the triangle is EM, which in this case is the vertical side. However, an is
Rollout 9
So that suggests something's wrong.\n\nWait, perhaps my assumption is incorrect. Let's recast this.
Rollout 2
5. Wait, in this case,\n\nbut solving the system leads x=1000n +
Rollout 8
as the permutations of the actual cards.\n\nBut maybe in terms of generating the happy sequences:\n\n- For each
Rollout 11
some combinations of them are real. Perhaps we can use some relationships here. Let's think about using the
Top 16 Negative Activations
Rollout 3
^2 = 28,224. 169^2 = 28
Rollout 3
root.\n\nsqrt(28,436). Let's see, 168^2
Rollout 2
k>=3 are10^k -1. For10^3=1000
Rollout 3
^2 = 28,561. So sqrt(28,436)
Rollout 10
Similarly, n=17≡2 mod5. Check moves from 17:\n\n17-
Rollout 4
001) is 0.01. Let's approximate:\n\n0.0000
Rollout 7
2=14/3 +3.5=14/3 +7/2= (
Rollout 10
15 is congruent to 0 mod5. Let's check:\n\nn=15: subtract
Rollout 10
another modulus.\n\nSuppose n=7 was losing. Check if n=14≡0 mod7
Rollout 7
. However, the problem states N<40. 36 is less than 40,
Rollout 13
about $9,441.12. The difference between these two amounts is how much more
Rollout 4
000 = 0.005. And for voltage V, ΔV / V =
Rollout 12
\), multiplied by the derivative of the upper limit. Let's compute \( b'(\nu) \
Rollout 1
2} approx 0.707d. So sum of radii l_s = d /
Rollout 2
For k>=3:2^k mod8. But 2^3=8≡0 mod
Rollout 10
move from 20 leads to a winning position. The moves from 20 are subtracting squares

DOWN_PROJ

Top 16 Positive Activations
Rollout 5
areas, and then find the greatest integer not exceeding x2/100. \n\nFirst, let
Rollout 4
, since the energy is proportional to C_eq * V2, the relative error in energy will be a
Rollout 1
_x, v_{x'}⟩ = (2/d2)⟨x - y, x' -
Rollout 3
73 K as STP, but I'm not sure. Alternatively, in the context of an example
Rollout 3
divided by (kg/mol) is (m2/s2). Square root is m/s. So unit
Rollout 12
) = \frac{1}{\sqrt{(x^2 - 1)(1 - \nu
Rollout 11
1 and q=3. But no, that seems arbitrary.\n\nWait, let me test with a specific
Rollout 11
= d, b = d, c =1/d2 as before, but complex d. For instance
Rollout 11
/(a b) = (1/(a b))2 / (a b) )= 1
Rollout 4
says "percentage error", and depending on how precise they want, maybe two significant figures? 0.
Rollout 11
(so same modulus and argument), c =1/a2. Then compute p and q. Test if
Rollout 10
thing in number theory, but I don't see a direct connection here.\n\nAlternatively, since in the game
Rollout 1
would be sqrt( (d / sqrt(2))2 + (d / sqrt(2))2
Rollout 4
%.\n\nBut the options or the answer expected? Possibly they want an answer to two decimal places. For example
Rollout 1
∈ S, ||x - y||2 = d2 / 2.\n\nExpanding, that gives
Rollout 1
If we have ||x - y||2 = d2 / 2 for all x, then from
Top 16 Negative Activations
Rollout 7
average54, total=54*9=486. Total=186+
Rollout 9
(10,10), which is the vertical side on the right. But wait, the problem
Rollout 10
Legendre's three-square theorem, a number n is expressible as a sum of three squares if and
Rollout 11
itself. So for p to be real, p = conjugate(p). Similarly for q. So let
Rollout 8
31, n=126, which are coprime, so sum m +n=
Rollout 14
Wait, another idea is to use the Cauchy-Schwarz inequality in a smart way by introducing
Rollout 3
.314*298= about 19,820 (same as earlier
Rollout 8
1: \(2 \times 5! = 2 \times 120 = 2
Rollout 14
J$ and $K$.\n\nMaybe Cauchy-Schwarz on $I$? Let's
Rollout 11
example can satisfy these conditions, meaning this approach is dead end.\n\nThis is perplexing. So using symmetric
Rollout 5
= 125y\n\nBring all terms to left side:\n\ny^2 - 25
Rollout 9
0,10), which is the vertical side on the right. But wait, the problem says the
Rollout 15
ent if and only if all its eigenvalues (in the algebraic closure) are zero. Not sure
Rollout 0
b\) and \(a\) and \(b\) are coprime positive integers. The product \(a
Rollout 10
��0 mod4: subtract0 or1.\n\n subtract s≡0 or1 mod4, n
Rollout 13
.\n\nWait, that seems familiar but let me verify using another approach.\n\nSince 1.25 *

Layer 58

GATE_PROJ

Top 16 Positive Activations
Rollout 13
[\n1.12^2 = 1.2544 \quad \text{and
Rollout 13
\n1.12^2 = 1.12 * 1.12. Let
Rollout 6
Meyer's theorem equivalence between \( \textbf{P}/\text{poly} \) and sparse
Rollout 13
�� 6000 * 1.57351936 ��
Rollout 13
.1255)^2 = 1.267506 approximately? Wait,
Rollout 13
2544^2 \approx 1.57351936\n\
Rollout 14
≈(1.31)^3≈2.25. But 8 J K=8
Rollout 13
1.03^4 �� 1.12550881, therefore
Rollout 13
.12^4\):\n\[\n1.12^2 = 1.25
Rollout 13
):\n\[\n1.12^2 = 1.2544 \quad \text
Rollout 3
0.222 �� 3.1415926536
Rollout 13
^2 = 1.03*1.03 = 1.0609
Rollout 13
1.03^4 �� 1.12550881\n\nThen
Rollout 3
0.222 �� 3.1416 * 0.22
Rollout 13
\):\n\[\n1.12^2 = 1.2544 \quad \
Rollout 6
P/poly is established that a language is in P/poly if and only if it is polynomial-time
Top 16 Negative Activations
Rollout 11
to the sum of the variables plus the sum of their products two at a time. That form makes me
Rollout 1
would be sqrt( (d / sqrt(2))2 + (d / sqrt(2))2
Rollout 11
see if such examples exist.\n\nExample 1:\n\nLet’s take a, b, c such that a
Rollout 5
areas, and then find the greatest integer not exceeding x2/100. \n\nFirst, let
Rollout 9
+ (0 - y)^2] = sqrt[h^2 + y^2]\n\nSimilarly, GM
Rollout 11
two variables are reciprocals. For example, let’s suppose a = 1/b, so that
Rollout 2
109≡5 mod8.\n\nTherefore, equation becomes:\n\n(5m +5)≡5
Rollout 11
). \n\nLet me think of a different approach. Let me let a, b, c be three numbers
Rollout 1
parallelogram law says that ||u + v||2 + ||u - v||2 =
Rollout 9
the square is labeled AIME, moving around the square, then maybe A is the bottom-left, I
Rollout 11
, maybe using Vieta’s formula. If we consider a polynomial equation whose roots are a, b,
Rollout 11
+ c +ab+ bc+ ca. compute p2. Maybe that expands to something that can be
Rollout 5
8125 = 0\n\nWait, that leads to x2 = 1812
Rollout 2
0^k - 1).\n\nSo the product we are dealing with is P = (10^
Rollout 4
so 0.88%, or maybe the answer is exact at 0.87%.\n\nBut
Rollout 1
x, y⟩ + ||y||2 = d2 / 2.\n\nSimilarly for another x',

UP_PROJ

Top 16 Positive Activations
Rollout 7
56N -56P=56N +15P=66N\n\nSo
Rollout 11
5/4) e^{i2θ} - (15/4) e^{-i2
Rollout 2
999=999*10 +9, wait, perhaps 9999
Rollout 7
9P +79x\n\n→-3P +A x -79x=0\n\n
Rollout 2
1) \equiv -891 \equiv -16 \equiv 109 \mod
Rollout 7
/3\n\nThus:\n\nx(A -42) +14N=56N/3
Rollout 7
6N -56P=56N +15P=66N\n\nSo
Rollout 7
6N -56P = 56N +15P=66N\n\nSo
Rollout 7
part(a):\n\nA*x +42*N/3 -42x=56N/3
Rollout 7
213N -205N)/3=8N/3.\n\nHence:\n\n3
Rollout 11
q =a/b + b/(1/(ab)) + (1/(ab))/a = a/b
Rollout 7
ide by N (N≠0):\n\n-2 + (A -79)/12=0
Rollout 12
) ] K(k)\n\n= [ nu K(k) - E(k)/nu ] / (1 -
Rollout 12
(k) / (k(1 -k2)) ) ] - [ K(k) /k ]
Rollout 9
(-5/h)x + (50/h) +0? Wait, let's verify:\n\nFrom G
Rollout 11
/4) e^{i2θ} - (15/4) e^{-i2θ
Top 16 Negative Activations
Rollout 15
all lamps will be off after some time.\n\nWait, but I need to be careful here. The problem
Rollout 10
=2 is a losing position, so P-position.\n\nWait, contradicts my previous thought. Wait:\n\n
Rollout 15
(L2) is on, right? Wait:\n\nWait, for L3, neighbor is L2
Rollout 13
calculation gives around 1.5735?\n\nAlternatively, I could use a calculator step here,
Rollout 10
is a losing position, so P-position.\n\nWait, contradicts my previous thought. Wait:\n\nIf n
Rollout 2
satisfies the first congruence.\n\nWait, hmm, that seems contradictory. Wait, no, according to
Rollout 2
8 is5 which is correct. Hmm. Wait, am I missing something here?\n\nWait, in our
Rollout 2
, which satisfies the first congruence.\n\nWait, hmm, that seems contradictory. Wait, no,
Rollout 12
the term involving the upper limit might vanish or something.\n\nAlternatively, maybe we can substitute variables to absorb the
Rollout 15
2. Therefore, n=2 is not acceptable.\n\nHmm, but wait, is it possible that another
Rollout 15
? Wait, O for on, T for off.\n\nSo OOO becomes OTO at t=1
Rollout 9
, the horizontal distance from G to EM.\n\nWait, maybe visualizing coordinates would help. Let's place
Rollout 4
But error propagation using partial derivatives is more efficient here.\n\nHowever, let me verify to make sure.\n\nThe
Rollout 4
4%), so maybe a couple of decimal points. \n\nAlternatively, perhaps compute the maximum and minimum possible values
Rollout 0
20! but 7^3 does not.\n\nBut exponent of 7 is 2, which
Rollout 9
0,0) lies within the square? Yes.\n\nWait, but then the area of the triangle in

DOWN_PROJ

Top 16 Positive Activations
Rollout 3
98) ) �� sqrt( ~225,753 ) �� 4
Rollout 3
/0.697433≈26,034.21.\n\nsqrt
Rollout 3
/ 0.697 �� 26,049.\n\nsqrt(26
Rollout 3
/0.697433≈26,050\n\nsqrt(26
Rollout 1
where all the distances are equal, it's called an equidistant set. In Hilbert spaces,
Rollout 3
/0.08796≈225,298. Then sqrt(2
Rollout 1
space is compact. However, Hilbert spaces are only locally compact in finite dimensions, thus only in finite
Rollout 6
think I'm going in circles here. Given the time constraints, maybe for problem 3, since it
Rollout 1
-separable Hilbert spaces, you have uncountable orthonormal bases, while in separable only
Rollout 10
,3,...\n\nTherefore, the losing positions are numbers of the form 5k and 5k +
Rollout 7
repeaters, N = P + R.\n\nWe are given:\n\n1. The average of all participants is
Rollout 1
bert spaces, you have uncountable orthonormal bases, while in separable only countable ones
Rollout 1
typically in non-separable Hilbert spaces, you have uncountable orthonormal bases, while in
Rollout 1
the dimension is the cardinality of any orthonormal basis. But H here is infinite-dimensional, which
Rollout 3
406, then sqrt≈168.5 m/s. So insignificant change.\n\nAlternatively,
Rollout 2
125*(8n)+109=1000n +109\n\n
Top 16 Negative Activations
Rollout 3
but as per my knowledge cutoff is 2023, so if need to answer, in the
Rollout 4
00 /4975= 4. Therefore, C_eq_min=1190
Rollout 3
2 * 270 = 17,958.24, plus 6
Rollout 11
+1/r3*1/r\n\n= r^4 + r2 +r^{-2} +
Rollout 7
leading to a max average of64.\n\nThis is impossible, meaning that assuming the given conditions(after increase
Rollout 11
itself. So for p to be real, p = conjugate(p). Similarly for q. So let
Rollout 7
scored 64). So A=98 is impossible.\n\nTherefore, inconsistency. What does that mean
Rollout 11
ab) didn't work. Trying a specific example seems not possible as it resulted in prohibited things. H
Rollout 7
2 +56=98.\n\nBut that's impossible as A must be <=64 (since
Rollout 8
the lengths of each block are variable (must be at least 1, obviously). But the key is
Rollout 8
, 31/126≈0.246.\n\nTo verify if that's correct
Rollout 13
.251 + 0.01575025 = 1.2
Rollout 15
". So, for n=4, there exists at least one initial configuration that leads to an infinite cycle
Rollout 15
, an important paper by Sutner states that linear cellular automata with next state being the sum mod
Rollout 4
00 + 3000) = (6,000,000)/
Rollout 4
sqrt(7.7) is approx 2.7749, so sqrt(7.

Layer 59

GATE_PROJ

Top 16 Positive Activations
Rollout 7
66N +5N =71N.\n\nBut separately, we can say total promoted score
Rollout 7
59(R - x) = 71N.\n\nAdditionally, we have original relations:\n\nOriginal total
Rollout 7
) = 76P + 61R = 66N +5N =7
Rollout 12
- k2 cos2 theta = 1 - k2 + k2 sintheta2 = (1
Rollout 12
is equivalent to 1 / sqrt(1 - k2 + k2 sintheta2 ). Hmm,
Rollout 12
- nu2)(1 - sin2θ ) = 1 - k2 cos2θ\n\nWait
Rollout 5
50))/2 * (h/2) = (2b + 50)/2
Rollout 7
for part(b) is N=24?\n\nWait, wait. Earlier we saw that in equations,
Rollout 1
c - d2/2) ) = 0 ⇒ (2c - d2)/2 -
Rollout 12
So this expression inside the sqrt is 1 - k2 cos2theta.\n\nBut then, f(n
Rollout 0
pairs (a,b) would be 2^8 = 256. However, because
Rollout 12
theta2 = (1 - k2) + k2 sin theta^2. Which is similar to
Rollout 1
parallelogram law says that ||u + v||2 + ||u - v||2 =
Rollout 11
1/(ab)) + (1/(ab))/a. Simplify each term: a/b is the
Rollout 12
we can set �� such that t2 = ν2 + (1 - ν2) sin2
Rollout 2
=999≡-1 mod125.\n\nIn fact, as noticed before, 1
Top 16 Negative Activations
Rollout 1
convex. But since H is infinite-dimensional, it is a complete metric space. However, the intersection might
Rollout 1
\mathcal{H}$ be an infinite-dimensional Hilbert space, let $ d>0
Rollout 3
(298 K), average speed is 1155 km/h, which translates to ~
Rollout 3
°C (298 K), average speed is 1155 km/h, which translates to
Rollout 1
, hence the desired y exists.\n\nSince H is a complete metric space, and the family of spheres has
Rollout 1
thonormal system must be countable because the space is separable? Wait, but the problem states that
Rollout 3
.512.\n\nThen, 66.512 * 298 = let
Rollout 1
-empty, hence the desired y exists.\n\nSince H is a complete metric space, and the family of spheres
Rollout 1
point \( y \) is guaranteed by the properties of infinite-dimensional Hilbert spaces, ensuring the required or
Rollout 1
the finite intersection property is non-empty if the space is compact. However, Hilbert spaces are only locally
Rollout 1
the whole intersection is non-empty, provided the space is complete (which it is) and perhaps the family
Rollout 1
. But since H is infinite-dimensional, it is a complete metric space. However, the intersection might still
Rollout 1
a non-empty intersection. In infinite dimensions, balls are not compact hence there is no guarantee.\n\nHences
Rollout 1
(has the finite intersection property) and maybe in a weakly compact set. But in Hilbert spaces
Rollout 1
the space is compact. However, Hilbert spaces are only locally compact in finite dimensions, thus only in
Rollout 12
2x^2)}}\] \n(where the positive value of the square root is taken) is monoton

UP_PROJ

Top 16 Positive Activations
Rollout 11
, then r3 +1/r3 >2 ( as AM ≥ GM, minimum at r=1
Rollout 1
x'.\n\nLet me recall that in Hilbert spaces, if we have a set of vectors with equal norms
Rollout 1
for distinct x, x'.\n\nLet me recall that in Hilbert spaces, if we have a set of
Rollout 0
in 20!:\n\nFor prime 2: floor(20/2) + floor(
Rollout 8
4/3024=744÷24=31; 302
Rollout 4
72). sqrt(18.72)= approximately 4.326 pF.\n\n
Rollout 14
on the left and combined three on the right. Maybe a Hölder inequality? Since Hölder can relate
Rollout 1
ormal systems can be uncountable. However, in that case, the problem still requires that {v
Rollout 9
vertical, then the base is vertical. But triangles are usually thought of as having a horizontal base, but
Rollout 14
the left and combined three on the right. Maybe a Hölder inequality? Since Hölder can relate multiple
Rollout 1
of a simplex or something similar. For example, in finite dimensions, if you have points on a sphere
Rollout 8
126 divides by 2, 3, etc. 126 is 2*
Rollout 5
, the midline divides the trapezoid into two smaller trapezoids? Each with height
Rollout 10
as sum of an even number of squares. Wait, sum of squares theorem is a thing in number theory
Rollout 1
zero for distinct x, x'.\n\nLet me recall that in Hilbert spaces, if we have a set
Rollout 1
where each pair is orthonormal. However, in Hilbert spaces, the dimension is the cardinality
Top 16 Negative Activations
Rollout 6
with α_n | n ∈ N }, then this set is sparse.\n\nTo retrieve α_n, the oracle
Rollout 11
imodular, so t1 + 1/t1 is complex. Hence, contradiction.\n\nSo this
Rollout 10
infinite set of P-positions. Thus, the set S is infinite.\n\nHence, this shows there
Rollout 14
2λ), and zero elsewhere.\n\nCompute the integrals:\n\nJ = ��0^{x_max}
Rollout 4
series. The voltage applied across this combination is $V=5.00 \pm 0.
Rollout 14
≤ (a^p)/p +(b^q)/q\n\nBut there are three terms here.
Rollout 1
distance from 0 to an arbitrary z in S' is ||z|| = ||x - x0
Rollout 6
be the binary encoding of α_n. But α_n has length p(n), so s_n is a
Rollout 10
set of P-positions. Thus, the set S is infinite.\n\nHence, this shows there are
Rollout 14
). The left-hand side is the cube of the integral of f. The right-hand side is 8
Rollout 8
5,5−r).\n\nSo:\n\nTotal happy sequences = sum over r=0 to 5 [
Rollout 6
= { α_n | n ∈ N }, each string is of length p(n), hence the number of
Rollout 15
. But over GF(2), where states are vectors with entries in GF(2), and addition is
Rollout 1
such that { x - y : x ∈ S'} are orthogonal.\n\nWait, original problem: when translating
Rollout 1
if r1 + r2 >= l and |r1 - r2| <= l.\n\nIn our
Rollout 4
. The relative error here is ΔC1 / C1 = 10 / 200

DOWN_PROJ

Top 16 Positive Activations
Rollout 3
terms of molar mass, since it's a monatomic gas, the molar mass would be approximately
Rollout 6
is similar to problem 1. If you consider countably infinite advice strings for all n, and represent
Rollout 15
.\n\nL4: neighbor L3 (off). Remains off.\n\nThus, t=4: O
Rollout 10
2;15,17;20,22;...\n\nSo each group is 0
Rollout 10
W.\n\nn=24: n -1=23 (W);n -4=2
Rollout 7
71N -205N/3=(213N -205N)/
Rollout 0
.\n\nTherefore, the previous answer 128 must be correct, hence The answer would be 1
Rollout 1
, ||x - y|| = d / sqrt(2) for all x in S. So every
Rollout 1
the vector x - y and x' - y must be orthogonal. That suggests that the inner product is
Rollout 7
/3 +7/2= (28 +21)/6=49/6≈
Rollout 7
)/12=2→A-79=24→A=103.\n\nBut
Rollout 10
mex{0,1}=2→Grundy[24}=2\n\nn=25:
Rollout 3
the isotope, then its molar mass is 222 g/mol, or 0.
Rollout 3
. At STP, the molar volume is 22.4 liters, which is for ideal
Rollout 7
24 works.\n\nN=36:\n\nP=24, R=12.\n\nx=
Rollout 15
automata have the property that every configuration evolves to the zero configuration.\n\nBased on the above, if the
Top 16 Negative Activations
Rollout 7
average54, total=54*9=486. Total=186+
Rollout 9
= sqrt[(10 - (10 - h))^2 + (0 - y)^2]
Rollout 12
(1 - ν2 - t2 + ν2)/(1 - ν2) ) = (1
Rollout 11
term:\n\na/b = (r e^{iθ})/( r e^{-iθ} )) =
Rollout 9
:\n\nThe slope (m) is (0 -5)/(10 - (10 - h))
Rollout 3
8,436). Let's see, 168^2 = 28,
Rollout 14
x_max, f(x)= (1 - μ x)/(2λ), and zero elsewhere.\n\nCompute the
Rollout 4
2 being C22/(C1 + C2)^2 and C12/(C1 +
Rollout 11
^{iθ}\n- c = 1/(r2 e^{i2θ})\n- ab =
Rollout 1
Hilbert space is istit's isomorphic to l2(I) for some index set I. Therefore
Rollout 13
1255. Alternatively, perhaps use the standard multiplication grid. But this might take longer.\n\nAlternatively
Rollout 12
sqrt( (x2 -1)(1 - ν2x2) ). Let me set a substitution
Rollout 5
0\n\nThis is a quadratic in y: y2 -150y -125
Rollout 11
= re^{iθ}, c =1/(r2 e^{i2θ}). Product a*b
Rollout 12
becomes sqrt( ( (1 - ν2)/ν2 ) * 0 ) = sqrt(0
Rollout 3
3,476.2) �� 183 m/s. So that's the

Layer 60

GATE_PROJ

Top 16 Positive Activations
Rollout 5
zoid into two regions whose areas are in the ratio $2: 3$ . Let $x
Rollout 15
". Therefore, the lamp's own state at t doesn't matter. The next state depends only on the
Rollout 5
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nOne base
Rollout 15
on the above, if the cellular automaton with rule based on binary XOR of neighboring states (as this
Rollout 1
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet
Rollout 4
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nTwo capac
Rollout 3
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nCompute the
Rollout 7
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nIn a
Rollout 9
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nSquare $
Rollout 6
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet's
Rollout 10
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nConsider the
Rollout 8
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nKathy
Rollout 12
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nProve
Rollout 11
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nSuppose
Rollout 14
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nA function
Rollout 15
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet
Top 16 Negative Activations
Rollout 14
):\n\nλ2 λ=1/12→λ3=1/12→λ=(1
Rollout 1
= 2d / sqrt(2) = d sqrt(2) > d, so the spheres
Rollout 11
1. Then abc= �� k * ��2 k *1/(k2 ��3
Rollout 12
t2)\n\nSo denominator becomes sqrt( ( (t2 - ν2)/ν2 ) (1
Rollout 11
= (1/a)/a2 = 1/a3\n\nTerm6: c2/(a b
Rollout 14
0 to1/μ: μ*(1/μ3)/3= 1/(3 μ2
Rollout 14
^3$.\n\nThus, LHS =1/k^3, RHS=4/k^3,
Rollout 14
, I^3/(J K)= (1/k^3)/ ( 8*(1/(2
Rollout 11
=k e^{-iθ}, thus, ab =k2. Then c=1/(ab)=1
Rollout 12
/ (1 - nu2 )\n\nBut since k = sqrt(1 - nu2) =i nu
Rollout 11
� k * ��2 k *1/(k2 ��3 )= ��^3
Rollout 12
's express 1 -k2. Since k = sqrt(1 - nu2), 1 -
Rollout 12
k2 =1 -nu2, so k = sqrt(1 - nu2) =: k
Rollout 11
line{a}\n\nThus, a/b = |b|^2 \overline{a} →Multiply
Rollout 12
1/ν of dx over sqrt( (x2 -1)(1 - ν2x2
Rollout 11
/(a^3)\n\nSo q =1 + a^3 + 1/a^3\n\nSince

UP_PROJ

Top 16 Positive Activations
Rollout 14
2 dx) * sqrt(∫12 dx) ) = sqrt(J) * sqrt(∫1
Rollout 14
} (�� x dx )^{3/2} ? But again, ��xdx diverges
Rollout 14
1^2 dx)(∫f^2 dx) by Cauchy-Schwarz, but as
Rollout 14
= sqrt(J) * sqrt(∫1 dx) = sqrt(1) *∞ =∞.
Rollout 5
In a trapezoid, the midline (the segment connecting the midpoints of the legs)
Rollout 1
space, any orthonormal system must be countable because the space is separable? Wait, but
Rollout 1
points in R^3 with equal pairwise distances d (i.e., an equilateral triangle in 3
Rollout 9
(length 10 -0=10) and height=10 (the distance between x
Rollout 1
there is something called a "pseudobasis" or "orthonormal basis" even if un
Rollout 5
same. But in general, trapezoids are not similar unless the ratio of the bases is the
Rollout 0
be \(2^8 = 256\), but each pair is counted twice except when \(
Rollout 14
A f2 dx ) * ��(A) by Cauchy-Schwarz.\n\nSimilarly,
Rollout 1
intersection property) and maybe in a weakly compact set. But in Hilbert spaces, closed balls are
Rollout 12
] dt = \frac{\pi}{2b} \n\nWait, this integral actually calculates a certain period
Rollout 14
1/x2 dx)^{1/2} by Cauchy-Schwarz.\n\nWait,
Rollout 4
) - a*b)/(a + b)^2) = (b(a + b - a)) /
Top 16 Negative Activations
Rollout 7
likely, the problem expects there to be solutions. So I must have made a mistake.\n\nWait, alternative
Rollout 7
. So I must have made a mistake.\n\nWait, alternative approach. Let me use another angle.\n\nGiven
Rollout 7
)\n - Substituting \( P = \frac{2N}{3} \) and \(
Rollout 7
given. But we have to check our steps.\n\nSo Part (b) asks: after 5 point
Rollout 13
6000 \left(1.12\right)^4\n\]\nCalculating \(
Rollout 10
L positions. That can't be true. Wait, that means after n=22, every position
Rollout 2
=> 0 mod1000, so10^4 -1=9999
Rollout 13
**\n\[\nA_{\text{annual}} = 6000 \left(1 +
Rollout 7
is impossible.\n\nWait, that can't be. So there must be a mistake here. Let me verify
Rollout 3
mann distribution:\n\n\[\nv = \sqrt{\frac{8RT}{\pi M}}\n\]\n\n
Rollout 2
000=>-891=109 modmod1000. Hence,
Rollout 13
4\n\]\nCalculating \(1.12^4\):\n\[\n1.12
Rollout 7
6 students average 54: total=54*6=324. Total=
Rollout 7
before. But how does this make sense?\n\nAlternatively, perhaps during the problem, the composition of promoted and
Rollout 4
F}$ and $C_{2}=3000 \pm 15 \mathrm{p
Rollout 2
=0 mod125, so 10^4 -1≡-1 mod12

DOWN_PROJ

Top 16 Positive Activations
Rollout 3
0.697 ��28,624; sqrt ��169
Rollout 4
327 / 12 = 0.360583333...
Rollout 3
/0.08796≈225,298. Then sqrt(2
Rollout 7
16 with total 71*16=1136. Increase by 5 each
Rollout 2
0 mod125\n\n10^3=1000 mod125.
Rollout 3
.697433≈28,622\n\nsqrt(28,6
Rollout 12
/ (1 - k2 sin2 theta)^{3/2} } ] d theta\n\nBut
Rollout 8
contiguous color blocks. Hence, such sequences are called "linear arrangements with two blocks".\n\nAlternatively, in comb
Rollout 7
x=2, sum=98*2=196. Original scores:avg=1
Rollout 3
98) ) �� sqrt( ~225,753 ) �� 4
Rollout 12
) ] d theta\n\ndK/dk = ��0^{pi/2} [ (k
Rollout 7
average62, total=62*3=186. Remaining 9 with average5
Rollout 8
order matters, the number of sequences is the number of injective functions from 5 positions to 1
Rollout 13
628 - 9441 = 187\n\nSubtract cents: 2
Rollout 3
222, with a half-life of about 3.8 days. But in terms of m
Rollout 9
Wait, but in trapezoid area formula:\n\nAverage of the two bases * height.\n\nBut the
Top 16 Negative Activations
Rollout 14
some sort of inner product involving x f(x). Hmmm.\n\nWait, what if we make the substitution
Rollout 14
right-hand side is 8 times the integral of f squared times integral of x f. Given that integr
Rollout 1
balls are not compact hence there is no guarantee.\n\nHences perhaps requires another approach, using the structure of
Rollout 3
/0.697433≈27,950\n\nsqrt(27
Rollout 3
/0.697433≈27,950. sqrt(27
Rollout 3
016÷0.697≈27,950; sqrt(27
Rollout 12
0 * infty, which is more nuanced. Hmmm. Let's see.\n\nTake x =1
Rollout 14
2 space. Maybe try expanding on a basis. Hmmm...\n\nWait I just had another thought. Maybe
Rollout 11
2/(a b) = (1/(a b))2 / (a b) )=
Rollout 11
/x ) + ca (which is (1/(xy)) * x = 1/y )\n\nSo yes
Rollout 14
x f. Given that integrals of xf and f squared are finite.\n\nThe function is defined from[
Rollout 3
75 /2.828≈168, which is about right. So, the
Rollout 3
512 * 270 = 17,958.24, plus
Rollout 5
areas, and then find the greatest integer not exceeding x2/100. \n\nFirst, let
Rollout 12
( nu2 + (1 - nu2) sin2θ ) ]\n\nSo, f(nu)
Rollout 3
sqrt(27,950)=167.2 m/s.\n\nSo approximately 1

Layer 61

GATE_PROJ

Top 16 Positive Activations
Rollout 0
total pairs, so 256/2 =128. Hence 128 such
Rollout 11
k, b= omega2 k, c=1/(omega3k2 )=1/k2
Rollout 11
z ) +a2 z +1/a +1/(a z )\n\nGrouping terms:\n\n= [
Rollout 11
Third: a/b * a/c = a^2/(b c)\n\nSimilarly, b/c * b
Rollout 0
a < b hence 256/2=128. So confirms.\n\nBut let me
Rollout 11
In上述 p: p= t1 +1/t1 + t2 +1/t2 + t
Rollout 11
work. So if arranging a= k e^{iθ}, b=k e^{-iθ}, c
Rollout 8
.\n\nThus, the answer would be 31 +126=157.\n\nBut I
Rollout 7
the equation becomes:\n\n209N/3 +16x =71N.\n\nNow,
Rollout 2
�5 mod8\n\nCompute 125 mod8: 125 /8=
Rollout 11
, c = 1/(a*b ) =1/(a^2 z). Hence, c =
Rollout 11
c=1/(omega3k2 )=1/k2, sinceomega3=1. But
Rollout 11
ab))/a = a/b + ab2 +1/(a2 b ). So earlier same expression.\n\n
Rollout 11
+ [ (c/b^2 + a^2/(b c) + b^2/(a
Rollout 14
μ2)= λ/μ=1→λ=μ.\n\nHence λ=μ.\n\nThen substit
Rollout 0
×3...×20 is given by 2^{number of prime factors of 20!
Top 16 Negative Activations
Rollout 5
b + 25}{b + 75} = \frac{2}{3}\n\
Rollout 5
+ 25}{b + 75} = \frac{2}{3}\n\]\n
Rollout 9
0 - \frac{500}{h} = 80 \implies \frac{
Rollout 4
}{2000 + 3000} = \frac{60000
Rollout 9
00 - \frac{500}{h} = 80 \implies \frac
Rollout 8
Case 1: \(2 \times 5! = 2 \times 120 =
Rollout 5
frac{18125}{100} = 181.25\n
Rollout 5
{18125}{100} = 181.25\n\
Rollout 13
1}\right)^{1 \cdot 4} = 6000 \left(1
Rollout 12
This transforms the integral into:\n\n\[\nf(\nu) = \int_{\nu}^1
Rollout 1
langle x - y, x' - y \rangle = 0 \).\n\n2. **Key Observ
Rollout 2
25\n \]\n - Let \(x = 125m + 10
Rollout 4
00000}{5000} = 1200 \, \text
Rollout 4
2000 + 3000} = \frac{600000
Rollout 0
\) is \(\frac{256}{2} = 128\).\n\nThus,
Rollout 5
25 + 30625}{2}} = \sqrt{\frac{362

UP_PROJ

Top 16 Positive Activations
Rollout 5
stituting these values:\n\[\nx = \sqrt{\frac{75^2 + 1
Rollout 12
\( f(\nu) = \int_1^{\frac{1}{\nu}} \frac
Rollout 5
175^2}{2}} = \sqrt{\frac{5625 + 3
Rollout 3
oltzmann distribution:\n\n\[\nv = \sqrt{\frac{8RT}{\pi M}}\n
Rollout 7
total:79*(P + x)+47*(R -x)=71N.\n\nSame
Rollout 5
30625}{2}} = \sqrt{\frac{36250}{2
Rollout 7
N/3) +79x +47*(N/3) -47x=
Rollout 8
: \(2 \times 5! = 2 \times 120 = 24
Rollout 0
1\), each prime factor of 20! must be assigned entirely to either \(a\)
Rollout 8
).\n - Total for Case 1: \(2 \times 5! = 2 \times
Rollout 14
-Schwarz inequality for \( I_1 \):\n\n\[\nI_1 \leq \left
Rollout 0
in lowest terms is \(20!\) is \(\boxed{128}\).<|im_end|>
Rollout 7
9*(2N/3 +x)+47*(N/3 -x)=71N
Rollout 7
) are \(\boxed{12}\), \(\boxed{24}\), and \(\
Rollout 7
P +x), non-promoted total=47*(R -x). Then:\n\n79(P
Rollout 8
**\n - There are 2 such sequences: RRRRR and GGGGG.\n - Number
Top 16 Negative Activations
Rollout 2
izing each term, might use modulus1000. However, since 1000=
Rollout 13
3^8:\n\nAs calculated prior step, but alternatively, compute 1.03^8 as
Rollout 13
, actually, if I check:\n\nLet me use a different approach. Since we had 1.0
Rollout 13
5735?\n\nAlternatively, I could use a calculator step here, but as I need to compute
Rollout 3
sqrt(8*R*T/(π*M))\n\nLet's calculate the numerator and denominator step by step.\n\n8
Rollout 11
be complex or real, but p is real, then (a+1)(b+1)(c
Rollout 13
36\n\nWait, or perhaps it's easier to multiply directly. Still, this might take time.\n\n
Rollout 10
manually spot a pattern here, but maybe there's a different approach.\n\nThe problem statement says that there are
Rollout 2
. But wait when doing straight mod1000 I also arrived at109, which is
Rollout 10
3, 2, 3. Is there a pattern? Let me check n=11.\n\n
Rollout 13
.5735?\n\nAlternatively, I could use a calculator step here, but as I need to
Rollout 3
mol unit cancels, then kg in denominator; thus overall (kg·m2)/(s2·
Rollout 2
5 is9 *99 * (-1). So, 9*99=891
Rollout 7
haven't considered.\n\nWait, perhaps non-integer averages? But average A=98 in original scores
Rollout 13
here, but as I need to compute manually... Maybe another way.\n\nAlternatively, write 1.2
Rollout 2
=3 is9*99*(-1). So from the previous step when we had9*9

DOWN_PROJ

Top 16 Positive Activations
Rollout 6
log n + n^k �� m? Unlikely. For example, m is dominated by n
Rollout 11
\overline{ab} + \overline{bc} + \overline{ca} =
Rollout 11
= a + b + c, T = ab + bc + ca. Then P = S + T
Rollout 11
involve terms like a/b +ab2 +1/(a2 b ). But not sure.\n\nAlternatively,
Rollout 14
_a *||x^(1/3)||_b,\n\nwith 1/a +1/b=
Rollout 1
d2 ) / 2 = 0.\n\nRearranging:\n\n||y||2 - ⟨
Rollout 11
= a +b +c, t=ab + bc + ca, but here p =s +
Rollout 11
. Suppose that a +b+ c +ab + bc + ca =-1 and a/b +
Rollout 11
text{and} & $q=\dfrac ab+\dfrac bc+\dfrac ca$
Rollout 14
\|f\|_a \|1\|_b,\n\nBut as before, if you use
Rollout 11
? p = a + b + c +ab+ bc+ ca. compute p2. Maybe that
Rollout 2
09 mod8: 109-13*8=109-10
Rollout 1
inner products.\n\nIf it's possible to find y such that translating by y makes the Gram matrix orthogonal,
Rollout 11
))/a = a/b + ab2 +1/(a2 b ). So earlier same expression.\n\nIs
Rollout 14
2<= sqrt(J1 ) sqrt A + K/(A).\n\nNow, notice that J1<=∫
Rollout 11
(a +b +c) + (ab + bc + ca) + abc=1. But
Top 16 Negative Activations
Rollout 0
20! or a < sqrt(20!). Therefore, the number of such a is half
Rollout 8
5 -k) = 5! / (k)!.\n\nSo the number for each color sequence (
Rollout 14
sqrt{J}} \right)^{2/3}}.\n\]\n\nSimplifying both terms,
Rollout 13
your answer as a dollar value to the nearest cent.\n<|im_start|>assistant\n<|im_start|>think\nOkay, so Jo
Rollout 1
vectors in $ \mathcal{H}$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I
Rollout 4
of the energy stored in this combination of capacitors?\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I
Rollout 14
= \frac{2K}{\sqrt{J}} \implies A = \left( \
Rollout 1
x' - y is zero for distinct x, x'.\n\nLet me recall that in Hilbert spaces,
Rollout 15
guarantee that all lamps will be off after some time?\n<|im_start|>assistant\n<|im_start|>think\nAlright, let's
Rollout 2
mod125.\n\nSince (-1)^{997}= -1.\n\nTherefore, 1
Rollout 10
second player can win no matter how his opponent plays.\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I
Rollout 12
- \nu^2)(1 - t^2)}}\n\]\n\nNext, we recognize this integral
Rollout 14
infty xf(x) dx \right) $$ \n<|im_start|>assistant\n<|im_start|>think\nOkay, so I
Rollout 4
008774964*100=0.87749
Rollout 5
2(b + 75); 3b + 75 = 2b + 1
Rollout 8
5 - k)!] * [5! / k!].\n\nThen, since for each k (from

Layer 62

GATE_PROJ

Top 16 Positive Activations
Rollout 8
$5$ of the cards in a row in a random order. She will be happy if and only
Rollout 4
^2}{(C_1 + C_2)^2} \quad \text{and}
Rollout 4
C_1 + C_2}\n\]\n\nGiven \( C_1 = 2000
Rollout 5
b = 175 \). Substituting these values:\n\[\nx = \sqrt{\
Rollout 5
{75^2 + 175^2}{2}} = \sqrt{\frac{
Rollout 8
. She shuffles the $10$ cards and lays out $5$ of the cards in a
Rollout 4
^2}{(C_1 + C_2)^2}\n\]\n\nSubstituting the values
Rollout 4
+ (0.16 \cdot 15)^2} = \sqrt{3.6
Rollout 8
order. She will be happy if and only if all the red cards laid out are adjacent and all the
Rollout 4
C_2)^2}\n\]\n\nSubstituting the values:\n\n\[\n\frac{\partial C
Rollout 8
will be happy if and only if all the red cards laid out are adjacent and all the green cards laid
Top 16 Negative Activations
Rollout 2
: 9 *99 * (-1)^{997} mod125\n\nBecause
Rollout 2
, product becomes9*99*(-1)^{997} mod1000.
Rollout 14
that a*2 +b(k +1 ) =1? Not sure.\n\nLet me attempt:\n\nSup
Rollout 14
||f||2 ||1||2 ||1||infty, but i think triple Hölder not
Rollout 6
$L$ . [/*]\n<|im_start|>assistant\n<|im_start|>think\nOkay, let me try to work
Rollout 11
θ - (1/8) sin3θ = (63/8) sin3θ\n\n
Rollout 1
balls are not compact hence there is no guarantee.\n\nHences perhaps requires another approach, using the structure of
Rollout 14
say 1/p +1/q +1/r=1. If I can split the integrand into
Rollout 11
+a)(1 +b)(1 +c)=1. Hence, (1 +a)(1
Rollout 14
c^6/(18 d^3 )=8/18*c^6/d^3
Rollout 1
= d sqrt(2} ~1.414 d > distances between centers d, hence intersection
Rollout 7
. Average=1350/18=75. Correct.\n\nNon-promoted:6
Rollout 12
4 ) ) =pi/2( k2/4 ) >0. So E(k) >
Rollout 1
over the Hilbert space, so thinking as in R^n, for which we know that even for three
Rollout 7
4.\n\nOriginal promoted P=16, R=8.\n\nAfter increase promoted total:79*(
Rollout 1
sqrt(2) �� 1.414d >= d, so intersection is non-empty

UP_PROJ

Top 16 Positive Activations
Rollout 1
:\n\n-2〈x. This action would be the same as above for linear equations, but we already
Rollout 1
formal differences. Imagine that the set S, translated by y, becomes orthonormal, rescaled.
Rollout 15
by Sutner states that linear cellular automata with next state being the sum mod2 of neighbor states
Rollout 6
, which might not be helpful.\n\nAlternatively, recall that P_angel is equivalent to P/poly,
Rollout 6
polynomial-time machine with a sparse oracle. (This is the Meyer, I think, theorem.) So perhaps
Rollout 6
be arbitrary. It is known that P/poly contains undecidable languages (as the advice can be
Rollout 6
, you can have for each n, a set of strings that encode the advice α_n. But since
Rollout 13
.03)^4 -1 = 1.12550881 -1
Rollout 14
helpful. \n\nWait, have you consider applying the Cauchy-Schwarz inequality in a form
Rollout 11
θ}, b=k e^{-iθ}, thus, ab =k2. Then c=1/(
Rollout 15
wait contradiction.\n\nWait in Sutners paper “Linear Cellular Automata and the Garden-of-Eden
Rollout 11
^2 - 2t or similar.\n\nBut in our cubic, s= a +b +c
Rollout 6
a polynomial-length advice string, which can be arbitrary. It is known that P/poly contains undec
Rollout 7
, after the error, all scores are increased by 5, so effectively, the new scores are original
Rollout 6
and only if it is polynomial-time Turing reducible to a sparse set. So that's the main
Rollout 6
the advice strings being generated non-uniformly. The class P/poly is equivalent to languages decidable
Top 16 Negative Activations
Rollout 14
\frac{f(x)}{\sqrt{x f(x)}}}$, but that feels like a stretch and
Rollout 14
infty xf(x) dx \right) $$ \n<|im_start|>assistant\n<|im_start|>think\nOkay, so I
Rollout 11
q equal to the conjugate of another term:\n\nSuppose a/b = \overline{(b/c
Rollout 14
c - d x for 0 ≤x ≤ c/d, 0 elsewhere.\n\nFirst, check that
Rollout 12
-E(k)/( (1 - nu2) nu )\n\nSecond term: [ nu2 / ( (
Rollout 11
, making s=1/r >1, f(r) = f(1/s)= s^{-3
Rollout 2
by $1000$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I have this problem
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) ] = dθ / t\n\nSo, rearranged, the integrand dt / sqrt(...) becomes
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{}!$ be the resulting product?\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I need to figure
Rollout 6
|} can be computed in poly-time given |x|, which is the length of x, then
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=1? Not sure.\n\nLet me attempt:\n\nSuppose we have:\n\nI = ��0^
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N/3\n\nTherefore:\n\nx(A -54) =56N/3 -18
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, |b|=k>1, then |c|=1/k2 <1. So none have
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A +5)*x] / (P +x)=79.\n\nSimilarly, total non-promoted
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’s call that approach. Let’s try that.\n\nSuppose we want to maximize $I^3$
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-5)/(10 - (10 - h)) = (-5)/h.\n\nThe equation is

DOWN_PROJ

Top 16 Positive Activations
Rollout 5
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 0
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 1
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 3
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 4
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 9
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 6
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 14
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 12
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 7
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 11
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 8
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 15
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 10
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 2
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 13
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Top 16 Negative Activations
Rollout 12
\nu^2)(1 - t^2)}}\n\]\n\nNext, we recognize this integral as
Rollout 11
work. Trying a specific example seems not possible as it resulted in prohibited things. Hmmm.\n\nWait,
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- t)(t + a)(t + b)} } \n\nBut maybe that's diverging.\n\nAlternatively
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, b, c replaced by their conjugates. Hmmm. To have q real, need x/y
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1/4. Which matches the ratio here. Hmmm.\n\nWait a minute, perhaps I made a
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becomes zero. Wait, over GF(2), it might actually require more steps, but somehow in our
Rollout 1
why in the previous example with translated set S', it encountered a problem?\n\nAh, because in that example
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1)} = K^{1/(a -2)}.\n\nTake logarithms:\n\n(1/(2a
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, 2^(3 -1)=4, which matches what we saw for 720.\n\n
Rollout 4
% error. \n\nBut since combining them by quadrature gives sqrt(0.52 +0.
Rollout 1
for a set S which includes 0, then it would require that 〈x, x'〉
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that the original rational number between 0 and 1 is written in reduced form, and you multiply numerator
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0]\n\nAnd cycles. Hence, the cycles' behavior differs. InGF(2) terms reaches zero
Rollout 8
R1 G1) has both red and green cards, each as singleton blocks. Are singleton blocks considered
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:\n\nWait, if over GF(2), the system is [1,1,1] -> [
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0 and other points. Then, the act of translation introduced coordinates that conflicts with the required inner products.

Layer 63

GATE_PROJ

Top 16 Positive Activations
Rollout 14
bit tricky. Maybe express this as:\n\nRearranged,\n\n$$\nf(x) = a -
Rollout 8
be adjacent). But the way the problem is phrased: "all the red cards laid out are
Rollout 8
k) = 5! / (5 - k)!.\n\nSimilarly, arrange (5 -k)
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. So the count is half the number of coprime ordered pairs, or is it?\n\nWait,
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and had three prime factors. The number of coprime pairs a < b where a*b=7
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\neq b\). Hence, total coprime pairs are \(2^{8}\), where
Rollout 5
connecting the midpoints of the legs divides the trapezoid into two regions with areas in the ratio
Rollout 0
, but when we write a and b as coprime factors, the way I'm thinking is that
Rollout 0
(2,3,5), number of coprime pairs where a < b is 4 (
Rollout 3
πM)), whereas the root mean square speed is sqrt(3RT/M). Since the question refers to
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is parallel to the bases and that divides the trapezoid into two regions of equal area. Find
Rollout 0
assignments will satisfy a < b, even for coprime divisors. Wait, but in reality,
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}, where d is some variable. But likely overcomplicating.\n\nAlternatively, assume a = d,
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sum of two squares, but I might be overcomplicating.\n\nAlternatively, looking back to the problem
Rollout 7
$75$ and that of the non-promoted $59$ .\n(a) Find
Rollout 2
be ��0 mod8. Therefore, m=8n for some integer n.\n\nTherefore, x
Top 16 Negative Activations
Rollout 13
is $12\%$. She makes no payments for 4 years, but has to pay
Rollout 3
\) is the molar mass of radon in kg/mol.\n\n**Steps:**\n\n1. **
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athy has $5$ red cards and $5$ green cards. She shuffles the $10
Rollout 0
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\n
Rollout 1
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\n
Rollout 4
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\n
Rollout 3
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\n
Rollout 11
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\n
Rollout 7
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\n
Rollout 9
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\n
Rollout 6
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\n
Rollout 15
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\n
Rollout 14
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\n
Rollout 8
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\n
Rollout 12
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\n
Rollout 10
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\n

UP_PROJ

Top 16 Positive Activations
Rollout 8
k) = 5! / (5 - k)!.\n\nSimilarly, arrange (5 -k)
Rollout 8
(n,k) = n! / (n - k)!.\n\nSimilarly, for a different color sequence with
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0*(1 + 0.12/4)^(4*4)\n\nFirst, calculate
Rollout 4
.327 / 1200 �� 0.003606
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6666... /1200 �� 0.003605
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0,150 /4975 �� Compute 4975*11
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0*(1 + 0.12/1)^(1*4)\nSimplify the numbers
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C(5,5−r)*(5−r)! = 5!/( (5 - r
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1.57351936 �� 9,441.11
Rollout 4
000076996) �� 0.008775
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87386*1.03 �� 1.229873
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>=3, hence (10^k -1)≡-1 mod125.\n\nThus
Rollout 8
(5,5−r) *(5−r)! (choosing 5−r greens and
Rollout 13
0.03689622 �� 1.266770
Rollout 3
76. sqrt(33,476)=183 m/s. But since the
Rollout 8
= 10! / (10-5)! = 10 * 9 *
Top 16 Negative Activations
Rollout 6
oracle can contain those positions. So, for each (n, i), where i is a bit position
Rollout 10
**:\n - Players take turns removing square numbers (1, 4, 9, 1
Rollout 10
positions) where every move leads to a winning position (N-position) for the opponent.\n\n1. **
Rollout 6
.\n\nAnother angle: Let S_L contain the tuple <n, α_n >. But encode this in
Rollout 3
ounding to Significant Figures:**\n Given the inputs (molar mass \(222 \, \
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bbles. The key is to identify losing positions (P-positions) where every move leads to a
Rollout 6
the *angel string* is $\textbf{not}$ similar to a *witness*
Rollout 6
n, create entries in S_L with the form (n, i, b) where b is the
Rollout 6
build a sparse oracle S_L consisting of all tuples <n, i, b>, where the i-th
Rollout 6
the length $n$ . Note that the *angel string* is $\textbf{not
Rollout 6
can be generated in polynomial time given the input length (i.e., 1^n), then the advice
Rollout 6
a unique identifier, perhaps the query is a pair (n, i) where i is a bit position
Rollout 8
counts, and divide by the total number of permutations (10P5).\n\nLet me try this approach
Rollout 3
**\n\n1. **Molar Mass of Radon (Rn):**\n Radon has a m
Rollout 3
molar mass of radon in kg/mol.\n\n**Steps:**\n\n1. **Molar Mass of
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represents losing positions and W represents winning positions.\n\nInitialize:\n\nn=0: L\n\nFor n >=1

DOWN_PROJ

Top 16 Positive Activations
Rollout 14
g(y). Then J scales with d^2 ��g^2 dx= d^2/c
Rollout 9
$ units. Isosceles triangle $GEM$ has base $EM$ , and the
Rollout 14
(�� f(x) dx )^3 = �� f(x) dx * �� f(y
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��0^∞ f(x) dx = ��0^∞ f(x) \cdot1
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�g^2 dx= d^2/c ��g(y)^2 dy. K or S
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related to given integrals? We have J = �� f2 dx and K=∫xf dx
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(t) dt=K.\n\nSo we have that ��0^∞ F(x) dx=K
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the form:\n\n(integholder's inequality as: ��f(x) dx = �� f(x
Rollout 14
a -1} I.\n\nJ_λ = ��0^∞ f_λ(x)^2
Rollout 14
] * [x],\n- [J] = ��f2 dx ~ [f]^2 *
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1.\n\nThus, Hölder gives:\n\nI = ��f(x)x^{1/2} *
Rollout 14
with Cauchy-Schwarz:\n\nI = ��0^∞ f(x) dx = �
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such that:\n\n∫f(x) dx ≤ ( ��f(x)^ 2dx )^{a
Rollout 14
^{4/3} *3^{2/3}/4 = (2^{4/3}/
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(x) to be some function where we can relate ��f dx to ��x f dx through
Rollout 14
(x), then integral of f scales as μ λ ��f dx. While the other integrals:\n\n
Top 16 Negative Activations
Rollout 11
} = 1/(\overline{ab})). Therefore, maybe expressions can be written in terms
Rollout 11
^2 \overline{a} + \overline{c}^2 \overline{
Rollout 11
me compose that:\n\nFrom conjugate equation: \overline{a} = |b|^2 a
Rollout 11
absolute value $1$ . Define\n\begin{tabular}{c c c} $p
Rollout 14
bit tricky. Maybe express this as:\n\nRearranged,\n\n$$\nf(x) = a -
Rollout 1
) / 2 = 0.\n\nRearranging:\n\n||y||2 - ⟨x +
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]\n\nSimplifying this expression, we find:\n\n\[\nf'(\nu) = -\frac
Rollout 7
-promoted $47$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I'm trying to
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b = \frac{\mu}{2 \lambda}$. So $f(x) = a - b
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} + \overline{bc} + \overline{ca} = x + y + z
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but depending on which primes. But here is the crux: instead of trying to group, this is
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\mathbb{N} \mapsto \mathbb{N}$ such that for each
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a )2 + ( (a2)/(a + b)^2 * Δb )2 )\n\nLet
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0.02 \mathrm{~V}$. What is the percentage error in the calculation of
Rollout 1
a point $ y\in\mathcal{H}$ such that \n\[ \left\
Rollout 5
^2/100$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. I